Tfairself tYU.:; ■• ' .. p. - c I TEACH YOURSELF BOOKS ALGEBRA Some other Teach Yourself Books Arithmetic Atomic Physics Banking Book-keeping Calculus Further Calculus Commercial Arithmetic Computer Programming Costing Data Processing Dynamics Economics Electronic Computers Engineering Science Geometry Insurance Investment Mathematics New Mathematics Money Operational Research Ready Reckoner Ready Reckoner (Decimal currency edition) The Slide Rule Speed Mathematics Simplified Statics and Hydrostatics Statistics Trigonometry I TEACH YOU RSELF ALGEBRA P. ABBOTT, B.A. TEACH YOURSELF BOOKS ST. PAUL'S HOUSE WARWICK LANE LONDON EC4 First printed 1942 This edition 1969 This edition copyright © 1969 The English Universities Press Ltd. All rights reserved. No part of thi3 publication may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopy, recording, or any information storage and retrieval system, without permission in writ- ing from the publisher. SBN 310 03501 4 Printed in Great Britain fur The English Universttitt Prut, Ud^ 67 Richard Clay (/A/ Chaucer Pros), Ltd., Bungay. Suffolk INTRODUCTION Algebra is such a wide and comprehensive subject that this volume cannot be regarded as anything more than an elementary intro- duction to it. It is an endeavour to enable the private student to learn something of the principles and foundations of the subject, thus enabling him to proceed to the study of more detailed and advanced treatises. It also provides, within the necessarily pre- scribed limits of such a book, that knowledge of Algebra which is required by a student of allied branches of Mathematics or in applications of Mathematics to Engineering, etc. Consequently some of those elementary sections of the subject which are of little use for these purposes have not been included. The exercises are progressive and designed both to enable the student to test his knowledge of the work he has studied and also to provide material for his training in that power of manipulation which is so essential. They contain few of the more complicated or academic problems which are beyond the practical requirements of the ordinary student. An Appendix contains, without exercises, a very brief summary of the meaning of Permutations and Combinations, the Binomial Theorem, and the nature of the roots of a Quadratic Equation, together with those formulae which students may require when beginning work on the Calculus or other branches of Mathematics. While the fundamental laws of Algebra have not been entirely overlooked, rigid proofs of them have been omitted, owing to exigencies of space. It is hoped, however, that the logical basis of the subject has not been seriously impaired by the omissions. Some emphasis has been placed on the graphical aspects of parts of the subject, since experience has shown that they prove stimu- lating and provide revealing help to the student. No previous mathematical knowledge is required for this work, beyond that of Arithmetic. References have occasionally been made to theorems in Geometry or Trigonometry for the benefit of those students who have some knowledge of them. The Author is desirous of expressing his indebtedness to Mr. C. E. Kerridge, B.Sc, for the use of a number of examples from National Certificate Mathematics, Vol. I, and also to Mr. H. Marshall, B.Sc., for the use of examples from Vol. II of the same work. He also desires to record his gratitude to Mr. S. R. Morrell for the valuable assistance he has given in the correction of proofs. P. Abbott. PUBLISHER'S NOTE TO 1969 EDITION This edition has been revised to include examples in U.K. decimal currency. The examples are therefore also valid for non-sterling decimal currencies. CONTENTS CHAPTER I THE MEANING OF ALGEBRA PAR*. 1. Algebra and Arithmetic 3. A Formula . pack I PAKA. PAGE 13 I 3. Transformation of a Formula. 16 13 7. Algebraic Forma . . . 1» CHAPTER II ELEMENTARY OPERATIONS 9 Symbols ol Operation . '.'3 17. Multiplication . , 18 10. Algebraic Expression. Terms 13 18. rowers of Numbers IS 11 Bracket* 33 IS. Multiplication of Powers 30 n. Coefficient . 34 11. Division of Powers 3'J 13. Addition and Subtraction M 33. Easy Fractions 3i 16. Evaluation by Substitution . 36 CHAPTER III BRACKETS AND OPERATIONS WITH THEM 35. Removal of Brackets . "I * Systems of Brackets 16. Addition and Subtraction ot Expressions witliin Brackets 40 | CHAPTER IV POSITIVE AND NEGATIVE NUMBERS wltb 30. The Scale of a Thermometer . S3. Motion in Opixxlte Directions 84. Positive and Negative Num- bers 37. Operations Numbers Negative 41. 43. Rules for Signs Powers 31 66 66 CHAPTER V SIMPLE EQUATIONS 44. Meaning of an Equation 46. Solving an Equation . 69 69 t7. Problems Leading to Simple Equations a, n. Treatment of Formulae Transformation of Formulae . CHAPTER VI FORMULAE 69 I 63. Literal Equations 76 71 Solution of Equations CHAPTER VII SIMULTANEOUS EQUATIONS Simultaneous I 68. Problems Leading to Slmul- 78 | taneous Equations . CONTENTS CHAPTER VIII GRAPHICAL REPRESENTATION OF QUANTITIES PARA. PACK r-ARA. PAGE 60. The Object of Graphical Work 88 64. Examples ol Grains aud their 61. The Column Graph . . 88 Uses .... 91 63. A Straight-lino Graph . 89 CHAPTER IX THE LAW OF A STRAIGHT LINE; CO-ORDINATES 63. The Law Represented by a 73. Equation of a Mraight Line Straight-line Graph . . 101 Passing Through the Origin HI 69. Graph of an Equation of the 74. Equation of a Straight Line First Degree . . .108 Not Passing Through the 71. Position In a Plane; Co- Origin 113 1 ordinates .... 100 76. Graphic Solution of Simul- 73. A Straight Line as a Locus . 109 taneous Equations 114 CHAPTER X MULTIPLICATION OF ALGEBRAICAL EXPRESSIONS 77. Product of Binomial Expres- 1 83. Product of Sum and Diilerence siona . . . us I CHAPTER XI FACTORS 116 87. Binomial Factors . . . 1SU 1 91. Factors of the Difference of Two Squares . 136 CHAPTER XII FRACTIONS 09. Laws of Fractions . Mil IDS. Addition and Subtraction 143 100. Reduction ol Fractions . M0 103. Simple Equations Involving 101. Multiplication and Division . 141 Fractions . . 146 CHAPTER XIII GRAPHS OF QUADRATIC FUNCTIONS 104. Constants and Variables 148 109. The Curve ol y - «■ 161 106. Dependent and Independent Variables . . .149 110. The Curve of y —- x' 193 111. The Curves of »- ax' . 164 106. Functions . . . .149 113. The Curves of y - *• ± m 166 107. Graph of a Function . .160 113. Change of Axis . . 168 108. Graph of a Function of Second 116. The Graph of y — *■ - 3z - 3 169 Degree . . . .161 119. The Graph of y - 12 - x — x> 161 CHAPTER XIV QUADRATIC EQUATIONS 110. Algebraic Solution . 166 139. Problems Leading to Quadra- 134. Solution by Factorisation . 170 tics . . . . 176 116. General Formula . . .173 130. Simultaneous Quadratics 178 PAJU. 134. Law? of Indices 1SS. Bxtatuton ol the Meaning of all In'ipi .... 184. iso. 180. CONTENTS CHAPTER XV INDICES PACE PARA. PACK . 184 138. Fractional Indices . . 189 140. Negative Indices . . .180 187 141. Standard Tonus ol Numbers . 193 CHAPTER XVI LOGARITHMS 144. A System of Logarithms . 197 146. Characteristic of a Logarithm 199 147. Mantissa of a Logarithm 199 Rules for the Dae of Loga- rithms 301 Change of Base of a System of Logarithms . . 300 CHAPTER XVII RATIO AND PROPORTION Meaning of a Ratio Proportion . . 313 314 187. Theorems on Ratio and Pro- portion .... 169. Constant Ratios . 318 US CHAPTER XVIII VARIATION 181. Direct Variation 220 1«8. To Plod the Law Connecting Two Variables . .333 173. Inverse Variation . 330 17S. Functions ol More than One Variable . ... 33*. Plotting Against a Number CHAPTER XIX THE DETERMINATION OF LAWS Power of I 181. Use of Logarithms 241 CHAPTER XX RATIONAL AND IRRATIONAL NUMBERS; SURDS 184. Irrational Numbers 249 | ISC. Operations with Surds . CHAPTER XXI ARITHMETICAL AND GEOMETRICAL SERIES 344 380 187. The Meaning of a Series 189. Arithmetic Progression . 194. Harmonic Progression . 165. Geometric Progression . 301. Infinite Geometric Series 255 396 260 361 366 307. Simple and Compound In- terest .... 373) 308. Accumulated Value of Periodi- cal Payments . .374 309. Annuities . . .374 APPENDIX Permutations and Combina- tions . . .377 The Binomial Theorem 279 The Roots of liquation . Quadratic 38* Answers Logarithm Tables 381 304 CHAPTER I THE MEANING OF ALGEBRA I. Algebra and Arithmetic. Algebra, like Arithmetic, deals with numbers. Funda- mentally the two subjects have much in common ; indeed, Algebra has been called " generalised Arithmetic ", though this is a very incomplete description of it. It would perhaps be more correct to say that Algebra is an extension of Arithmetic. Both subjects employ the fundamental operations of addition, subtraction, multiplication and division of numbers, subject to the same laws. In each the same symbols, +, — , x , -f-, are used to indicate these operations, but in Algebra, as new processes are developed, new symbols are invented to assist the operations. Terms such as frac- tions, ratio, proportion, square root, etc., have the same meaning in both subjects, and the same rules govern their use. In Arithmetic we employ definite numbers; we operate with these and obtain definite numerical results. Whereas in Algebra, while we may use definite numbers on occa- sions, we are, in the main, concerned with general expres- sion and general results, in which letters or other symbols represent numbers not named or specified. This may seem vague to a beginner, but the following simple example may serve to show what is meant. 2. A formula. In Arithmetic we leam that to find the area of the floor of a room, rectangular in shape, we " multiply the length of the room by the breadth ". This might be expressed in the form : The area of a rectangle in square feet is equal to the length in feet multiplied by the breadth In feet. '3 , 4 TEACH YOURSELF ALGEBRA This rule is shortened in Algebra by employing letters as symbols, to represent the quantities. Thus : Let the letter / represent " the length in feet ". „ b „ " the breadth in feet ". " „ A „ " the area in sq. feet ". With these symbols the above rule can now be written in the form : A =1 xb. In this shortened form we express the rule for finding the area of any rectangle; it is a general rule, and is called a formula. The full description of it would be " The formula for finding the area of a rectangle ". It will be noticed that in the formula above there is no mention of units. This is because it is true whatever units are employed. It is necessary, however, that the same kind of unit should be employed throughout. If I and b are measured in inches, A will be the area in square inches ; if they are measured in feet, A will be in square feet. When using the formula for any specific case it is im- portant to state clearly what unit is employed for / and 6. The unit for A will then follow. Algebra has been called a kind of shorthand, and the above example in which a sentence has been reduced to A =1 X b illustrates the reason for the description, but as progress is made the student will discover that it is an incomplete description. From such a simple beginning as the above the subject develops into the most powerful instrument employed in Mathematics. The Greeks produced some of the greatest mathema- ticians in history, but their work was mainly accomplished in Geometry. They made very little progress in Algebra, since the symbols they could employ were extremely few. They did not even possess separate symbols for numbers, such as were afterwards introduced through the Arabs, and which we use to-day, but used instead the letters of the alphabet to represent numbers. The Romans were similarly restricted, and neither nation employed the decimal system of notation. The student will realise THE MEANING OF ALGEBRA ■5 something of the great difference suitable symbols make in Mathematics if he will, as an example, write down the number of the year in Roman numerals and then try to multiply it, say by 18, also expressed in Roman fashion. For progress in manipulation and general development of the subject not only do we require symbols, but they must be suitable for the purpose. The history of Mathe- matics reveals that many of the symbols which are so familiar to us now have been reached by a slow evolutionary process, often lasting through centuries. When letters are used to represent numbers any suitable choice may be made. In the above formula, although A, I and b were employed, any other letters could be used if they were more suitable. By common usage, however, certain letters are usually selected for specific purposes, and the same symbols are used invariably to denote certain numbers, as will become apparent to the student as he progresses. But whatever letters are employed in solving problems, the student must carefully observe the following: (1) It must be clearly seated what each symbol represents. (2) If measurements of any kind are Involved, the unit employed must be clearly defined. 3. Transformation of a formula. The student may be inclined to wonder why such an elementary rule as the area of a rectangle should be expressed as a formula by the employment of algebraic symbols. But this is an example selected for its simplicity to illus- trate the meaning of an algebraic formula. Even in this case it is possible in some measure to illustrate the flexibility and adaptability of a formula as compared with a statement in words. For example, suppose that I and b were measured in inches, then A = I x b sq. ins. If it is required to express this result in square feet we could do it thus: a I x b «, 4— jjj.aq.ft 16 TEACH YOURSELF ALGEBRA If the result were required in square yards, then: Or suppose that the rectangle represents a room, /and 6 being measured in feet, and that the room is covered with carpet costing a pence per square yard. If the cost of the whole carpet be represented by C pence, then: Since then ,4 = < J- 6 sq. yds. 7 x b C = — q X a pence. If this were to be expressed in pounds, then: — 9 x 100* Thus a formula may be used as a foundation for other formulae to express modifications of the original. It may also be transformed and used in various ways for other purposes. Suppose, for example, that the area and the length are known. Then if the breadth is required, it can be found by dividing the area by the length. This expressed as a formula would be : 4. An Illustration from numbers. The following example illustrates the use of letters to represent generalisations in number. We know that : (1) If any integer — i.e., a whole number — be multi- plied: by 2, the result is always an even number. (2) If any even number be increased by unity, the result is an odd number. These two statements can be combined in one as follows: If any Integer be multiplied by two and the product Increased by unity, the result Is an odd number. This is a generalisation about an odd number, expressed THE MEANING OF ALGEBRA «7 in words. This can be expressed by means of algebraical symbols as follows: (1) Let n be any integer. (2) Then 2 x n is always an even number. (3) Therefore 2 X » + 1 is always an odd number. In (3) we have reached an algebraical expression by means of which any odd number can be represented. The brevity and lucidity of this expression as compared with the full description of an odd number above will be apparent. But its value goes beyond this. We can manipulate this algebraic form, we can operate with it, and so can use it in the solution of problems. We may first note, however, that when expressing the product of two or more numbers represented by letters or a numeral and letters, the sign of multiplication can be omitted. Thus 2 x « can be written as 2n, and 2 X n + 1 as 2» + 1. This cannot be done with two numerals, such as 25, because under the decimal system the figure 2 has a place value. Multiplication may also be shown by a dot. thus 2 . n. If any odd number can be represented by 2n + 1, then, since when any odd or even number is increased by 2 the result is the next odd or even number, therefore the next odd number greater than 2w + 1 is In + I -f 2 or 2n + 3. Similarly 2n + 5 is the next odd number above 2« + 3. Consequently the expression 2n + 1, 2» + 3, 2» + 5, 2n + 7 . . . etc., represent a series, or a succession of consecutive Increasing odd numbers. Similarly, if an even number be diminished by 1, we obtain an odd number. .'. 2n — 1, 2» — 3, 2« — 5 ... , represent a series of decreasing odd numbers. Note. — The succession of " dots " after the sets of odd and even numbers indicates that we could write down more such numbers if it were necessary. 5. Substitution. In the algebraic representation of a set of odd numbers — viz.: In + 1, 2n + 3, 2n + 5 . . . i8 TEACH YOURSELF ALGEBRA since M represents any integer, we could, by assigning some particular value to ft, obtain the corresponding odd number. THE MEANING OF ALGEBRA 19 Thus if Then Similarly, n = 50. 2n + 1 = 2 X 50 + 1 = 100+1 = 101. 2« + 3=2x50 + 3 = 100 + 3 = 103. Consequently the series of increasing odd numbers corre- sponding to this particular value of n is: 101, 103, 105, 107 .. . Similarly, the decreasing odd numbers when m = 50 can be found by substituting this value of n in the series: 2n — 1, 2n — 3, In — 5 . . . Then we get the corresponding arithmetical values: 99, 97. 95 . . . 6. Letters represent numbers, not quantities. Such things as length, weight, cost, are called quantities. In short, anything which can be measured is called a quantity. , _, Letters are not used to represent these quantities. 1 nus in the formula which we stated above — viz. : A=lxb. I stands for the number of inches, or feet or yards, as the case may be, and to obtain it we must first determine the unit to measure the quantity, and then use a letter to represent the number of units. It is important, therefore, that when letters are to be used in algebraical expressions, it should be clearly stated what each letter represents. Thus we should say : Let I represent the number of inches in the length or, more briefly, let the length be I inches. Similarly let n represent the number of men, and let c represent the cost in shillings. We do sometimes, for brevity, and somewhat loosely, speak, for example, of the area of a rectangle as A, when we should say A sq. ft. or A units of area. The sign = may be used loosely in abbreviation of the statements above. Thus we could write: Let / = the length in inches, or let c = the cost in shillings. The sign = means " equals " or " is equal to ". It should connect two expressions which are equal in magni- tude, but is often used loosely as above to express equality. 7. Examples of algebraic forms. We now give a few examples of what may be termed algebraic forms — i.e., the expression in algebraic symbols and signs of operation of statements about quantities. Example I. Express in algebraic form the number of pence in x pounds added to y pence. To express pounds in pence we multiply by 100. .'. x pounds = 100 X x pence, and the total number of pence is lOOx + y. Example 2. A motor travels for t hours at v miles per hour. How far does it go ? How far will it go in 20 min. P The motor goes v miles in 1 hr. ,*. 2v miles in 2 hrs., 3v miles in 3 hrs., and in t hrs. it must go t x v miles. Since 20 min. is J hr., and the motor travels v miles in lhr. y .'. in J hr. it travels = miles. Example 3. There are two numbers; the first is multi- plied by 3 and 5 is added to the product. This sum is divided oy 4 times Hie second number. Express the result in algebraic form. We must begin by choosing letters to represent the unknown number. Let x represent the first number. „ y represent the second number. Then three times * increased by 5 will be represented by 3x + 5, and four times the second number is 4y. The division of the first expression by this is — -j — . 20 TEACH YOURSELF ALGEBRA Exercise I. 1. Write down expressions for: (1) The number of pence in £x. (2) „ „ pounds in n pence. 2. If £a be divided among n boys, how many pence will 63.ch bov £6t? 3 If n men subscribe £a each and m other men sub- scribe b pence each, how many pence are subscribed in all? 4. Write down the number of 1) Yards in a miles. 2) Miles in x feet. 3) Tons in c cwt. 6. The sum of two numbers is 28. If one number be n, what is the other ? 6. The difference between two numbers is x; if one ol them is 50, what is the other? ej . . 7. The product of two numbers is a and one of tnem is x; what is the other? 8. If the average length of a man s step is x inches, (1) How many yards will he walk in 100 steps? (2) How many steps will he take in walking a mile? 0. A number (*) is multiplied by 2, and 5 is added to the sum. Write down an expression for the result. 10. If x is an odd number write down expressions for 1) The next odd numbers above and below it. 2) The next even numbers above and below it. 11. A man buys sheep; x of them cost him a pence each and v of them cost b pence each. What was the total cost in pounds? 12. A number is represented by x; double it, add 6 to the result, and then divide the whole by 6y. Write down an expression for the result. 13. What number must be subtracted from a to give 6? 14. What number divided by x gives v as a quotient? 15. What is the number which exceeds 6 by a ? THE MEANING OF ALGEBRA 21 16. The numerator of a fraction is * increased by 2. The denominator is y diminished by 5. Write down the fraction. 17. A motor travels for m hours at p miles per hour. It then travels n hours at q miles per hour. How many miles does it travel in all ? 18. What is the total number of pence in £a + b pence. 19. A train travels at v miles per hour. How far does it go in x hours and how long does it take to go y miles? 20. Write down five consecutive whole numbers, the middle one of which is n. CHAPTER II ELEMENTARY OPERATIONS IN ALGEBRA 8. Use of symbols. In order that the student may become familiar with the processes of Algebra he needs considerable practice in the use of symbols. Consequently in this and subsequent chapters he will constantly be using letters which represent numbers in a general way and without any reference to quantities such as length, cost, etc., as was done in the previous chapter. Thus when we employ the form a + b we shall, in general, be using the letters, not as referring to any particular quantity, but as standing for any numbers. 9. Symbols of operation. As stated in § 1, certain symbols of operation, such as +, _, x, -f-, V. are common to Arithmetic and Algebra, since they are used for operations which are performed in both subjects. Usually, however, there is a certain difference in the way they are employed. It is evident that while such operations as 5 + 7, 10 - 3, 6 X 4, 15 -^ 3, V9 can be, and usually are carried out at once with definite numerical results, expressions such as a + b, a — b, a X b, a -f- b, Va cannot be evaluated numerically while a and b represent any numbers. Until numerical values are assigned to them we cannot proceed further with the operation. But we can, and do, operate with the expressions themselves, without any reference to their numerical values. In addition to the above, many other symbols of opera- tion are used in Algebra, among them the following: 32 ELEMENTARY OPERATIONS IN ALGEBRA 23 Symbol. Meaning. > < See § 6. is not equal to. is approximately equal to. is greater than, is less than. 10. Algebraic expression. Terms. Such a combination of letters and symbols as 2a + b is an example of what is called an algebraic expression. It may be defined as follows: An algebraic expression Is a combination of symbols which stand for numbers and for operations with them. For brevity the term " expression " is usually employed. When the expression contains the symbols of operation + or — , those parts of the expression which they separate arc called terms. Thus 2a -f 36 is an expression of two terms or a binomial. -k — y + 6* is an expression of three terms or a trinomial. A combination of letters which does not contain either of the signs + or — may be said to be an expression of one term, or a monomial. Thus -^- is such an expression. 11. Brackets. It frequently happens that an expression, or part of an expression, is to be operated as a whole. For example, suppose we wish to write in algebraic symbols " Twice the sum of a and b ". Evidently the arrangement 2 x a + b does not make it clear whether the 2 is to multiply a only or the sum of a and b. Consequently we employ " brackets " to enclose the part which is to be operated on as a whole — viz., a + b. :. we write 2(a + b). In this arrangement the multiplication sign is omitted between the 2 and the bracket. ^ TEACH YOURSELF ALGEBRA The brackets have the effect of indicating the order in which operations are to be carried out. Thus: 2(a + b) -c means that we find the sum of a and 6. multiply this by 2 and then subtract c. Similarly (a + b) X (c + d) or (a + b)(c + d) means that we find the sum of a and b. and also of c and d and then multiply the two results. This will be considered further at a later stage. 12. Coefficient. The expression 3a denotes a multiple of a and the number 3, which indicates the multiple, is called the coefficient of a. The coefficient may be a definite number like 3, called a numerical coefficient, or it may be a letter representing a number. , Thus in the expression ax, a may be regarded as the coefficient of x, but in some problems when we are thinking of multiples of a, x would be a coefficient of a ; in such a case we would usually write the expression as xa. In general, if an expression is the product of a number of factors, any one of them can be regarded as the coefficient of the product of the others, when for any purpose we regard this product as a separate number. Thus in 3ab, 3 is the coefficient of ab 3« „ ., b 36 „ .. a. Like terms. In an expression, terms which involve the same letter, and differ only in the coefficients of this letter, are called like terms. Thus in the expression 3a + 5b — 2a + 4b 3a and 2a are like terms, and 56 and 46 „ .. 13. Addition and subtraction of like terms. In Aritiimetic we learn that the sum oi 5 dozen and 9 dozen is 14 dozen, or (5 X 12) + (9 x 12) = 14 x 12 . . (A) ELEMENTARY OPERATIONS IN ALGEBRA 2> Similarly 8 score + 7 score = 15 score or (8 x 20) + (7 x 20) = 15 x 20. So for any number as for example (6 x 24) + (II x 24) = 17 x 24. In Algebra, if we were to let a represent 12 in the statement (A) given at the beginning of this section, we could write 5a + 9a = 14a, and for the other cases 8a + 7a = 15a and 6a + 11a = 17a. These last three cases are generalised forms of the pre- ceding examples, but it must be noted that whereas in the arithmetical forms we can proceed to calculate the actual value of the sum in each, in the algebraical forms we can proceed no further in the evaluation until a definite numerical value is assigned to a. Subtraction leads to similar results, just as so 9 dozen — 5 dozen = 4 dozen. 9a — oa = 4«. In this way we can add or subtract like terms only. It is not possible, for example, to perform any addition of two unlike terms such as 9a + 56. The rule for adding together like terms will now be clear. It is " add the coefficients ". Thus the sum of 2x + 5x + 3s = 10*. whatever x may be. The operation of " finding the sum " is used to include both addition and subtraction. This is called the " algebraic sum ". When an expression contains more than one set of like terms, these should be collected and dealt with separately. 14. Worked examples. Example I. Simplify 5a 4- 66 + 2a — 36. Collecting like terms 5a + 2a = 7a 66 — 36 = 36. Hence the whole expression is equal to 7a 4- 36. In 2 6 TEACH YOURSELF ALGEBRA practice there is generally no need to write down the above steps. The calculations can be made mentally. Example 2. Simplify 15* — 3y + 6y + lx — 5 - Collecting like terms and adding coefficients we get 22x + 3y - 5. 15. The order of addition. The counting of a number of things is not affected by the order in which they are counted. Thus : 6 apples + 4 apples is the same in number as 4 apples + 6 apples. This will be clear when it is remembered that 6 is the symbol for 6 units, and 4 is the symbol for 4 units. Thus: 6 + 4= (1 + 1 + 1+1 + 1 + 1) + (1 + 1 + 1 + 1) and 4 + 6 = (1 + 1 + 1 + 1) + (1 + 1 + 1 + 1 + 1 + !)• In each case the total number of units is the same. Thus algebraically 6a + 4a is the same in value as 4a + 6a. This is true for any algebraical sum. Thus 6a + 56 — 3a can be written as 56 — 3a + 6a without altering the value of the expression. Briefly the order in which numbers may be added is immaterial. 16. Evaluation by substitution. If we wish to find the numerical value of an algebraical expression for definite numerical values of the letters composing it, the expression should first be simplified by adding like terms. Then the numerical values are substi- tuted for the letters. Example. Find the value of 6* + 2y — 3* + 4y — 3 when x = 3 and y = 2. ELEMENTARY OPERATIONS IN ALGEBRA 27 Simplify the expression as in § 14 : 6* + 2y — 3x ; + 4y - 3 = 3x + 6y — 3. Substituting the given values : 3* + 6y - 3 = (3 x 3) + (6 x 2) - 3 = 9+12 — 3 = 18. Note. — ft will be seen that brackets are introduced when it is desirable to keep terms separate for evaluation. Exercise 2. 1 a) Find the value ot 6 dozen + 4 dozen. 6) Simplify 6a + 4a and find its value when a = 12. 2 a) Find the value of (8 X 73) - (3 x 73). 6) Find the simplest form of 86 — 36 and find its value when 6 = 73. 3. Write down in its simplest form: a+a+a+6+6+6+6 and find its value when a = 5 and 6 = 8. 4. Add together 2a, 4a, a, 5a, and 7a and find the value of the sum when a = 2-5. 5. Write the following expressions in their simplest forms : 15* — 3* + lx. 4x + 3* — 2* — x. (1) 156 + 116. (2) (3) 9a - 4a + 6a + a. (4) 6. Write in their simplest forms: (1) 5a - 26 - 3a + 66. (2) Up + 5o - 2o +p. (3) a - 2 + 36 + 6 + 5a. 7. Add together: 1) 4a - 56, a + 66, 5a + 6. 2) 6 + c — 3a", c + 26 + d, d — b — c. 3) 5x + 2y + 3z, * - y — 2z, 2x — y + z. 8. When a = 2, 6 = 3, find the numerical values of (1) 3a + 26 + 1. (3) 6a + 26 - 3a + 1. (2) 5a - 36 + 6. (4) 4a - 56 — 26 + 12a. 28 TEACH YOURSELF ALGEBRA 9. Simplify the following expressions and find their values when a = 4. b = 2, * : = 3, y = 5. 4ab — lab + Gab. 5ax — 2ax + bx. Gxy — ixy + xy. ab + Gbx — ay + 3a6 - 2bx. 10. Find the numerical value of a + \a + 1" + i a W 11. When' x = l.y = 2, z = 3, find the numerical value of 3* + 5y — 4* + 8* - y ■+ 5z. 12. When * = 4, y = 5. z = 1, find the values of (1) 3*y + 2yz — 8f - (3) ay + y + a: + 1. 1. (2) ay + yz + **• 13. If n be an odd number, write down the next three odd numbers greater than it and find their sum. 14 Write down a series of four numbers of which the first is a, and each of the others is twice the one which precedes it. Find their sum. 15. Write down a series of 5 numbers ol which the hrst is a, the second is greater than the first by d and each of the other three is greater by d than the one which precedes it. Find their sum. . 16. There are 5 numbers, the smallest of which is ex- pressed by 2m + 5. Each of the others is 3 greater than the one which precedes it. Write down the numbers and find their sum. 17. Multiplication. Order of multiplication. In Algebra, as in Arithmetic, the multiplication of a number of factors may be performed in any order, or, more precisely: The product of a number of factors is independent of me order in which tliey are multiplied. Thus 3 x 4 is equal in value to 4 X 3 0x3xo„ ., 3x5x6 ELEMENTARY OPERATIONS IN ALGEBRA 29 and generally a x b is equal in value to b x a and a x b x c „ .. c x b x a. Consequently il it is required to multiply say, 2a by 5 we can write the product thus: and 2xiix6 = 2x5x« = 10a 3ax 2b = 3x2xflx6 = 6a6. It ... Fig they 1(a). Fig. l(fc). would be arranged as in In this last example, a and b being unlike letters, we cannot proceed further with the multiplication. It should be noted, however, that although 4x3 is equal in value to 3 X 4, the two products do not necessarily mean the same thing when they refer to quantities. If, for example, 12 .... ... soldiers were to " form fours ", they would be arranged as shown in Fig. 1(a), the arrow showing the direction in which they are facing. But if the same 12 men were to " form threes Fig. 1(6). Thus 3 rows of 4 men require the same number of men as 4 rows of 3 men, but they are a different arrangement. Similarly if 4 men pay 3 shillings each ; the total amount paid is the same as that when 3 men pay 4 shillings each. 18. Powers of numbers. The product of equal numbers Is called a power. Thus: 8 x 8 is called the second power of 8, or the square of 8. 8 x 8 x 8 is called the third power of 8, or the cube of 8, there being three equal factors. 8x8x8x8 is called the fourth power of 8, there being four equal factors. Similarly: a x a is the second power of a or the square of a. a X a x a „ third „ a „ cube of a. a x a X a x a ,, fourth ,, a. 3„ TEACH YOURSELF ALGEBRA The process of writing a power in full is tedious, and the form of it restricts further operations, especially when the power is a high one. Accordingly mathematicians made many attempts through centuries to devise a symbolic method of representing the row of factors. Finally, Descartes in 1637 used a numeral to mark the number ot factors or the power and wrote the cube of a for example as a 8 , the fourth power as a 4 , etc. The figure used in this way is called an Index or exponent; it indicates the number of factors. Thus axaxaXaXais written as a s . With this symbolic method it is as easy to wnte down the 20th power of a number as the 2nd. But a new symbol, if it is to be satisfactory must not only express clearly and concisely the purpose for which it was devised, but it must also be convenient for operations with it. We shall see, from what follows, that an index fulfils this condition, and later in the book it will be seen how it lends itself to important developments. 19. Multiplication of powers of a number. Suppose we require to multiply two powers of a— say a* X a s . These numbers written in full are (a X a) X (a X a X a) the brackets serving to separate the two powers It is a fundamental law of Algebra, which will be assumed here, that when two groups of factors are multiplied, the factors in the groups are associated as one group of factors to give the product. By this law taxa)x(axaX a) = (a X a X a X a X a). • the number of factors in the product is the sum of the number of factors m the two groups. • in the example above the product (a X a X a x a X a) is the 5th power of a. and the Index of the product Is the sum of the Indices of the two factors. = a 5 . The same reasoning may be applied to other cases and so ELEMENTARY OPERATIONS IN ALGEBRA 3" we may deduce the general rule for the multiplication of two powers of a number. When two powers of the same number are multiplied, the index of the product is the sum of the indices of the factors. Examples. (1) x* X x* = x**' = x*. (2) 2a 7 x a s = 2a' * 3 = 2a 10 . (3) 56* X 36 8 = 5 x 3 X 6 8 X 6» = 15 X b* t8 - 156'. (4) a 2 b X ab* = a* X a X b X b* = a't 3 . The rule may be extended to the product of more than two factors. Thus: * ! X X s X x* = x*** + * = *». 20. Power of a product. To find the value of (ab)*. The use of the bracket shows that, as stated in § 11, the expression within the bracket must be regarded as a whole. ,'. by definition of a power ( fl 6) s = lab) x (ab) = (a X a) X (6 X b) = a*6*. Thus we see that the effect of this is that the index 2 must be distributed over each of the factors. Thus: (2xy)» = 2 3 X x 3 x y* = 8*y. So in Arithmetic (2 X 5)* = 2* X 6* = 4 x 25 = 100. Consequently when taking a power of a product, the index of the power is said to be distributed and applied to each factor of the product. Exercise 3. Write down the following in their simplest forms: 1. 4a X 3. 3. J* X 4y. 2. 5x x 2y. 4. 1m x 'in. 32 TEACH YOURSELF ALGEBRA 5. Ja X J6. 6. 6a X 56. 7. 3a x 46 X 5c. 9. ** x x. 11. x* X x\ 13. 2a 2 X a 8 . 15. 2a6 X ab. 17. x*y x *y*. 19. 2a X 3a» X a 8 . 21. (**)*. 23. (2a 8 ) 4 . Find the numerical values of the following: 25. 2a a x a, when a = 3. 26. 2a* + a, when a = 3. 27. a*6 x ab 2 , when a = 1, b = 2. 28. *• + 7* + 2, when * = 10. 29. 3a s + 2a6 + b*. when a = 2, 6 = 3. 30. c 3 X c* x c*. when c = 1. 31. 3a X 3a x 3a, when a = 2. 32. (2a i x) t , when a = 2, x = 3. 8. x y z 2 X 3 X I - 10. a X a* X a. 12. a 8 X a 8 . 14. 3s 8 X 2x*. 16. 26 X 34». 18. 7x*a X x*a. 20. (3a 2 6) 8 . 22. (2o*) 8 . 24. (4a) * x 4a» 21. Division of powers. Suppose that a power of a number is divided by another power of the same number, as for example, a 8 -7- a 8 . Every division can be expressed in fractional form, as in Arithmetic. a" 4- a s = a 55 _a xaxaxaxa a X a (by definition of a power). As in Arithmetic, common factors in the numerator and denominator can be cancelled. .'. the two a factors in the denominator can be cancelled with two of the five factors of the numerator. ELEMENTARY OPERATIONS IN ALGEBRA 33 Then there will be left in the numerator (5 — 2) factors. a 8 -T- a* = a 8 " 8 = a». Clearly the same method can be followed whatever the powers. Consequently we may deduce the rule: When dividing a power of a number by another power of the same number, subtract the index of the divisor from the Index of the dividend. Example. Divide 84a 6 by 12a 8 . Example. Divide 3x* by 6z e . Arranging as above. 3x* 3x* -f- 6x« = 6x* _3 x x x x XX x_x 6xxxxxxxxxxxx* In this example as the higher power is in the de- nominator, on cancelling there are (6 — 4) factors in the denominator. Hence we get: 1 1 2x*=* 2x v Exercise 4. Write down answers to the following: 1. a" -7- a. 2. a* H- a*. 3. 3x» -7- x*. 4. 6 s 4- 26. 5. 6a»-r-3a*. 6. 5y« -f-y. 7. 6*' -f- 2x*. 8. 14c* 4- 7c*. 9. a*^ -r- a6 s . 10. *V -r *V- 11. 5a»6» 4- a6*. 12. 6x*y -7- 2**. 13. 0a« -7- 3a«. 14. 15* 8 -7- 12**. 22. Easy fractions. Algebraic fractions obey the same fundamental laws as fractions in Arithmetic. In principle they are manipulated 34 TEACH YOURSELF ALGEBRA by the same methods. But since the numerators and de- nominators may be algebraical expressions, sometimes rather complicated, they present difficulties not found in Arith- metic fractions. In this chapter we shall deal only with simple forms involving easy manipulation, more difficult cases being left until later. 23. Addition and subtraction. The methods of Arithmetic can readily be applied, as shown in the following examples. Example I. Find the sum o/g + g- This is of the same form as 3 + 5 and is worked in the same way. T . 2.2 (2X6L+J2 X 3) justas g + g =^ $ft r so x x _ (*_ x 5) + (x x 3) 3 + 5~ 3x5 5x + 3* _ 8x ** 15 ~ IS - 3 4 Example 2. Find the sum of - + t- This is similar in type to the preceding and is dealt with in the same way: 3 4 _ (3 X 6) + (4 X a) a + 6" ~ ab _ 36 + 4o " ab~ • Example 3. Simplify - — t. Proceeding as before x _ a _ (x x b) — (a X y) y b ~ yx b bx — ay bf~' ELEMENTARY OPERATIONS IN ALGEBRA 35 Example 4. c . .,., 2« . 56 As in Arithmetic, we find the L.C.M. of the denominators —viz. 60. Then * =? ^2a , 56*. (2a X 4) + (56 x 5) i 15 "*" 12 i 00 _ 8o + 256 _ 60 - Note.— It is not possible to cancel any factors of 8 or 25 with factors of 60. This mistake is sometimes made by beginners, although the proceeding is contrary to the laws of Arithmetic. Only factors common to each term of the numerator can be cancelled with factors of the denominator. Example 5. y 18ab* Simplify jg^tj- We must find the L.C.M. of the denominators. To do this we find the L.C.M. of the numerical coefficients 12 and IS— i.e., 36— then the L.C.M. of a 2 6 and ab 2 . This is « 2 6 2 , since both of them will divide into it exactly. The product of 36 and a*6* is the L.C.M. of the denominators. _ (x x 36) — (y x 2a) 36« 2 6 a 12a 2 6 18a6 2 _ 36x — lay 36o 2 6~" 24. Multiplication and division. As in Arithmetic, these operations are based upon the same important rule of fractions, viz. : // numerator and denominator are divided by Die same number, the value of the fraction is unaltered. This can be expressed algebraically as follows: Let t be any fraction. Then M,d -FT* where m and n are any numbers. a V a V a X n b x n a-r-m 36 TEACH YOURSELF ALGEBRA 4s*y Example I. Simplify g^4. Written in full the fraction is: 4xsxsxsxy 'Qxxxyxyxy Cancelling common factors this is equal to: 2_x_x_x_x = 2x* 3 x y x y = 5y»' Note. — In practice there is no need to write out the powers in full as shown above. The rule for the division of powers may be applied directly. (Sax* 2y* Example 2. Simplify ^pp X g^t As in Arithmetic, factors in either numerator can be cancelled with factors in either denominator. 6ax«_ If 1555? X 3a* s-« 2>^J ~ X ?=* s » x 2y = 7 x a* = 7o»' 8s* 4s* Example 3. StM#./y ^ -f- -^. Proceeding as with a similar arithmetical example: 6a*y - 3a = 5a^ 4? i= 2x_x_3 r>«v _ 6x = 5 y- ELEMENTARY OPERATIONS IN ALGEBRA Exercise 5. Simplify the following: 37 1 * + ? 3.* + * * y _ 9s 5s lOy 8/ 5a_ _36_ '• Gftc^SaV 9. 3a6 + |=. ll.a-fl+1. 13 - i * + *- + -" ao oe ac 15. % xy loaWc "" lOaWc 5- 4s* x 2s»y 19 - 8s» • 25. 2s x -. 27.?-* y y 29 3fl - 9 -? *»• 7 • 14 - B (ALG.) 2s 3s 2 - T ~ T- 3a 5a *" 2b + 2b' 6 £ + £_£ 3y ^ 9y 6/ 10. 8 - .-5,. oy* 12. s - 1 + -. 14 * * + * a*6c «6 2 c a6c ! 1G. 18. 20. 12a 2 4 3a« • X 3 X X* *« a* X be* a 3 X b*c* 22. x* 3y» xy 2x 24. *" * £ 26. 2s-*. y 28. ?^.y y ' «" 30. 2s* . 3y 38 TEACH YOURSELF ALGEBRA 4a6c . 3c* dd - 5F ■ WaT>- 35. 30 -i . 3xyz 2z ' 32 «'-^ 34. Zxyz+^. 36 - ^SS" CHAPTER III BRACKETS AND OPERATIONS WITH THEM 25. Removal of brackets. Simple examples of the use of brackets as a convenient way of grouping numbers have already been considered. In this chapter we will examine extensions of their use, operations with them and the simplification of algebraical expression, which contain ._ _ ... l, brackets, by removing them. We will begin with a simple but important case. Fig. 2 represents a rectangle made up of two other rect- angles. Let a ins. = the length of one rectangle „ 6 ins. = „ „ the other rectangle. Let x ins. = the breadth of each rectangle as shown in Fig. 2. Ihcn (a + b) ins. = the length of the combined rect- angle, placing (a + 6) in brackets shown in § 11. The areas of two smaller rectangles are xa and xb sq. ins. and area of whole rectangle = x(a + b) sq. ins. But the area of the whole rectangle equals the sum of the area of the parts. x(a + b) = xa + xb. Similarly if there are three rectangles of lengths a. b and c ins. respectively, then in the same way we could show that: x{a + b + c) = xa + xb + xc. By modifications of the figures, which are left to the ingenuity of the student, we could similarly show that: (1) x(a -b) = xa — xb. (2) x(a + b — c) = xa + xb — xc. (3) x{a — b -f c) = xa — xb + xc. 39 4° TEACH YOURSELF ALGEBRA In all these examples expressions containing brackets have been transformed into expressions without brackets, or, as we say, the brackets have been removed. Hence we can deduce that: When the whole of an expression within brackets is multi- plied by any number, then, if the brackets be removed, each term within the brackets must be multiplied by the number. The factor without the brackets is said to be distributed as a factor of each term within the brackets. This is an example of the algebraical law called the Law of distribution. and subtraction of expressions within They are 26. Addition brackets. There are four cases which we will consider, represented by the following expressions: (1 (2 (3 4 a + a + a — a — b + c b-c b + c b-c The question to be considered is, What will be the effect of removing the brackets in the above expressions, a, b and c representing any numbers ? (1) a + (b + c) Let a, b, c be represented by the areas of rectangles as shown in Fig. 3. + ( b O a be Fig. 3. It is evident that the area of the whole rectangle which represents a + (b + c) is the sum of the three rectangles representing a, b, c. a + (b + c) = o + b + c. The steps in the addition of the numbers in this case are not altered by the insertion or the removal of the brackets. BRACKETS AND OPERATIONS WITH THEM 41 (2) a + (b-c) ^1 *■ 1>- Fig. 4. As in the previous case, a, b, c are represented by the areas of rectangles as shown in Fig. 4. The two unshaded rectangles represent a and b — c. .'. the whole unshaded portion represents a + (b — c). It can be formed either: (1) by adding (b — c) to a. (2) by adding b to a and then subtracting c. These two results are equal. + (b - c) = a + b - c. Thus no change results when the brackets are removed. <-- - Ml 1 Fig. 6. (3) a -(b + c) Using the same method as before of representing a, b and c, the unshaded rect- angle (Fig. 5) represents the remainder when (b + c), the two shaded rectangles, are subtracted from a, the whole rectangle, i.e., it represents a — (b + c). Also if from a, the whole rectangle, we subtract b and c in turn the remaining rectangle is the unshaded portion. .'. it represents a — b — c. a — {b + c) = — b — c. (4) a -(b-c) In Fig. 6 the rectangle representing a, b, c is shown, a being represented by ^ K _ the whole rectangle. The shaded rectangle represents (b — c). The unshaded rect- angle represents the result of subtracting (b — c) from a, i.e., it represents a — (b — c). m .--a—. Fig. 8. 4 2 TEACH YOURSELF ALGEBRA It may also be considered as representing the result of subtracting 6 from a and then adding c, i.e. it represents a — b + c. a — (b — c) = a — b + c. Collecting the four cases we have: (1 a + (b + c) = a + b + c. (2 a + (b — c) = a + b -c. (3 a — (b -f c) = a — b — c. (4 a — (b — c) = a - b + c. From these results we can deduce two rules respecting signs when the brackets are removed. A. From (1) and (2) when the + sign precedes the brackets the signs of the terms within the brackets are unaltered. B. From (3) and (4) when the — sign precedes the brackets the signs of the terms within the brackets are changed. 27. Worked examples. The following are examples of the use made of the rules of § 25 and § 20 when brackets are removed from an alge- braic expression in order to simplify it. Example I . Simplify a(a % + ab + 6 s ). When the brackets are removed the rule of § 25 is employed and the factor a multiplies each term within the brackets. Thus: a(a l + ab + 6 1 ) = a 3 + a*b + ab*. Example 2. Simplify 2(ia + 36) + 6(2a - b). When removing the brackets we use § 25 to multiply by 2 and fi, and from § 26, since the + sign before the second pair of brackets is positive, there is no change of sign. 2 (4a + 36) + 6(2a — b) = 8a + 66 + 12a — 66 = 20o. Example 3. Simplify 5x — (5y + 2x). This is an example of case (3) of § 2.6. On removing the brackets signs are changed. BRACKETS AND OPERATIONS WITH THEM O Thus: 6* - (5y + 2x) y-2x y (on adding like terms). = Bx — by — 2x = 3x - 5y Example 4. Simplify 3(4« — 6) — 2(3a — 26). This involves the rule of § 25 and case (4) of § 26. Using these 3(4a - 6) - 2(3a - 26) = 12a — 36 - 6a + 46 = 6a + b. Example 5. Simplify x(2x — y) — x(x — y) — y(x -f- 2y) and find its value wlien x=2,y = I. x(2x —y)— x(x — y)— y(x + 2y) = 2x z — xy — * 2 + xy — xy — 2y* = x % — xy — 2y % (since + xy — xy = 0). Substituting x = 2, y = 1. x t _ xy _ 2yt = (2)2 _ (2 x 1) - 2(1)» =4—2-2 = 0. Note. — This example shows the advantage of simplifying the expression before substituting the values of * and y. Exercise 6. Simplify the following expressions by removing brackets. < 3(5* + 6z). 1. 2. 2a(3a + 46). 3. 6a»(3a + 76 — 6c). 4. 2(* + 2y) + 3(2*-y). 5. x\x* — 3*) + x*(4x + 7). 6. i < x -2y) + l(2x + 4y). 7. 2(*+;y + 2)+3(2*+.y-2z). 8. * - (2y + z). 9. 2x — (y — 2z). 2(2a + 26) - 3(a - 6). 3a - (2a + 6). 3a — (2a - 6). 5x— (x — 2y + 2z). 3(„ + b - c) - 2(a - b + c). 10. 11. 12. 13. 14. 15. 4(x +y)-'3(2x-y) + 2(x - 2y). 44 TEACH YOURSELF ALGEBRA 16. a{a + b) — b{a — b). 17. X s (x + y) - xy{x* - y*). 18. 3(* s + * + 5) — 2(* 2 -3x — 4). 19. 2p(3p + -2q) - 3(7(2* - Sq) + p(3p 20. 5(*y) 2 - 3.r(y 21. (2% 2 ) 2 — 2x 3 (* 5?). *)• 28. Systems of brackets. It may happen that an expression within brackets is fart of another expression which is itself within brackets, n that case a second set of brackets would be required, and to avoid confusion they must be of a shape different from those already used, such as { } or [ ]. For example : 40 - {2(a -4- b) + 5(a - b)}. The student will easily recognise how clearly and effec- tively the brackets help' to show the construction of the expression and relations of the different parts to one another. It might happen that the whole of the above expression is to be multiplied by 2b. This will necessitate another set of brackets which will indicate that the expression is to be treated as a whole. We would express this as follows: 26[40 - {2(a + b) + 5(a - b)}]. Sometimes a straight line may be placed over a part of an expression with the same meaning as brackets. Thus x —y + z has the same meaning as x — (y + z). It must also be remembered that an expression which is the numerator of a fraction must be regarded as a whole and must be treated as if it were in a bracket. i-i_ _ a — 6 Thus: 5a ^ — means the same as 5a — J (a — b). If the whole expression were multiplied by 2, it would become 10a — (a — b). When expressions with two or more sets of brackets are to be simplified by removal of the brackets, it is well, as a rule, to begin with the innermost and work outwards. Examples of this will be seen in the following. BRACKETS AND OPERATIONS WITH THEM 45 29. Worked examples. Example I. Simplify 2{3a 4- 5(6 + c)}. As stated above we begin by removing the inner brackets. 2{3a + 5{b + c)} = 2{3a -f- 56 4- 5c} = 60 + 10b + 10c. Example 2. Simplify 3{3a - 2(a - 6)}. 3{3a — 2(a — 6)} = 3{3a — 2a + 26} = 3{a + 26} = 3o 4- 6b. Example 3. Simplify 12a - 2[3a — {4 - 2(a - 3)}]. Beginning with the innermost bracket 12a — 2[3a — {4 — 2 (a - 3)}] = 12a — 2 3a — {4 - 2a 4- 6}] = 12a — 2 3a — {10 - 2a}] (adding like terms) = 12a — 2 3a - 10 4- 2a] = 12a - 2 oa - 10] = 12a - 10a 4- 20 = la 4- 20. Exercise 7. Remove the brackets from the tollowing expressions and simplify them. 1. 3{5a — 3(a 4- 1)}. 2. 3{4(a 4- 6) - 3(a - 26)}. 3. tf6x - 3(2* - 1)}. 4. 5a 2 4- 2a{6 — (a 4- c)}. 5. 3/.*- {2^-^-1)}. 6. 3x(x 4- 3y) — 2{x s J- 3y(.r - 2y)}. 7. 36c - 2{6(6 - c) - c(b + c)\. 8. lox — [3* - {2* - (x — 5)}]. 9. 50 - 2[3a 4 - 2{36 - 4(6 - 1)]}. 10. 2(x+y) —x—y. 11.12{^-«-+ 6 4« TEACH YOURSELF ALGEBRA 12 . *{*^-*-^}. 13. 3c -{ 3 6 2a + c a + 2c S i*}- Fill in the blanks within the brackets in the following: 14. 2a — 6 + c = 2a — ( ). 15. x — y — 2 = x— ( ). 16. 2a + 46 — 6c = 2( ). 17. x 2 — xy+y* = x*-y{ ). 18. From 3a — 26 + 4c subtract a + 26 — 3c. 19. Take 2x — 2y + 4z from 3x — y + 2z. 20. When a = 3, 6 = 2, c = 1, find the values of the following: (1) 4a(a + 46) — a(3a — 6). (2) 3c{4c - (3c - 1)}. io=- 5" 5" • 0" CHAPTER IV POSITIVE AND NEGATIVE NUMBERS 30. The scale of a thermometer. Figs. 7a and 76 represent portions of Centigrade ther- mometers in which a fine column of mercury registers the rise and fall of temperature. The zero point, marked 0, indicates the position of the mercury in the tube at freezing point — i.e., the freezing point of water. Fig. 7(a) shows the mercury at 8° above zero. Now suppose the temperature falls 16° below this point. First it falls 8° to 0°, and then continues to fall for 8° below zero. To show this tem- perature on the scale it must be marked in q. some way which is different from the 8° above zero, or there would be confusion. To dis- tinguish the degrees below zero from those above we use the plan of putting a minus sign, — , before all those below zero, and if necessary a plus sign, -}-, before those above zero. Thus + 8° means 8 degrees above zero, and — 8" „ 8 „ below zero. We may call these positive and negative degrees, using the terms, and the signs + and — , to indicate different directions up and down from the zero. 31. An example from time. In reckoning time, the numbers denoting the years are counted from the birth of Christ. Years after that event are denoted by a.d., as a.d. 1941, and those before by B.C., as 55 B.C. The use of the symbols a.d. and B.C. is, in prin- 47 -8- -8- j -»• -10- ] I I ill <«) lb) Fig. 7. 48 TEACH YOURSELF ALGEBRA ciple, similar to the use of + and — in the case of the thermometer. 32. A commercial Illustration. In a certain transaction at market a farmer made a profit of £12. In a second deal he lost £8. Consequently in the two transactions lie made a net gain of £12 -£8 = £4. In a third transaction he lost £10. His profit and loss account is now shown by £4 — £10. If this loss had been £4, his position would be £4 — £4 = — i.e., he has reached a zero position, neither loss nor gain. But he lost £10, not £4; therefore he has a net loss of £6 — i.e., he is £6 below his zero. To distinguish gains from losses, we could, as in the case of the thermometer, place the negative sign before amounts showing losses, and the positive sign before amounts showing profits. In that sense £4 — £10 = — £6, the negative sign indicating a loss of £6. 33. Motion in opposite directions. Suppose a man starting from a point O {see Fig. 8) travels for 4 miles in the direction to X, reaching the point marked A. F E C •I »2 '3 •* »9 «6 -7 FlC. 8. He then turns and travels miles in the opposite direction, to X 1 . After 4 miles he reaches O, his zero or starting point. The next two miles take him to B. He is now 2 miles from O but in the direction opposite to that in which he started. His successive distances from O can be shown by +4 — 6. This suggests that, as in the previous cases, if distances from O in one direction were POSITIVE AND NEGATIVE NUMBERS 49 regarded as positive, the distances in the opposite direction could be regarded as negative. Thus, if we now say that the man is — 2 miles from O, we mean that he is 2 miles in the direction opposite to the original. Accordingly, in the diagram showing the movements from O (Fig. 8) Distances to the right with -f signs we would call positive. „ left with — signs „ ,, negative. With this device, when giving his position from O, the sign of the number would indicate in which direction the man is from O. Thus — 4 miles would indicate he is at C, + 2 miles would show he is at D. The number with the + sign we call a positive number. „ „ — sign „ negative number. 34. Positive and negative numbers. From this it appears that we have devised a new kind of number — viz., a negative number — and that, in conse- quence, we can divide numbers into two kinds: positive and negative. From the examples above a negative number Is a number which in its meaning and effect Is opposite to a positive number. Frequently, as in examples of §§ 30 and 33, the negative number indicates a direction opposite to that of the positive number, and in this sense, positive and negative numbers are called directed numbers. If negative numbers can rightly be classed as numbers, they must, in operations with them, conform to the rules governing the numbers which we now call positive numbers. These operations will be considered fully later, but a few simple illustrations will serve to show that we can deal with them in the same way as positive numbers. For example, in the matter of addition, we can add — 2 and — 3, and a glance at Fig. 8 will show that the result is — 5, being equal to the sum of — 2 and another — 3 from O to the point marked E. Or if — 3 be multiplied by 2 — i.e., we double the distance from O — we get — 6, at the point F. 5° TEACH YOURSELF ALGEBRA Similarly division of — 6 by 2 would give — 3. For the rest of this chapter, in order to make the meaning clear, positive and negative numbers, when being used in operations will be placed in brackets. Thus (- 6) -4- (+ 2) = (- 3). 35. Negative numbers. Corresponding to every positive number there is a negative number, and we can write a succession of negative numbers corresponding to positive numbers. Thus if we write down the numbers beginning, for ex- ample, with + 6 and decreasing by one at each step, we get the series of numbers + 6, + 5, + 4, + 3, + 2, + 1, With the negative number, we do not stop at the zero, but continue with the subtraction, so that we get — 1, — 2, — 3, — 4, — 5 . . . , etc. In descending order of magnitude. Or if we start with (—6) and add unity in succession we get the complete series : _ 6i _ 5 , _4, -3, -2, -1, +1. +2. +3, +4. +5 in ascending order of magnitude. Thisseriescan be extended in either direction and decimals and fractions fall into their places between these numbers. Thus we get what is called the complete number scale. 36. Graphic representation of the complete number scale. The graphical representation of the complete number scale is so important that we return to it again, using square ruled paper. gll i gag Tft tfgtfigff Up i lunMra * i ) ani.i5f:<-E:3.::-2IHAl.-r.O s ;1 . 4:* fct |* 3g& 1 Fig. 9. POSITIVE AND NEGATIVE NUMBERS 5> The straight line XOX' is drawn, as in Fig. 9, to show die representation of a small part of the scale. On this line, starting from a point O, and using a suitable scale, distances are marked to the right to represent positive numbers and to the left to represent negative numbers. We could imagine this line to be extended to any distance on either side so that any number could be included. Numbers involving decimals he between those marked. Thus — 2-5 would be at A. Two principles may be noted: (1) Every number can be represented at its appropriate point on the scale. (2) Conversely, every point on the scale represents a number. It should be observed that the numbers represented in the figure Increase from left to right, as shown by the arrow. 37. Operations with negative numbers. With the introduction of negative numbers, Algebra passes beyond the boundaries of Arithmetic. We must therefore proceed to examine operations with this new kind of number, remembering, as previously stated, that it must conform to the laws of Algebra. The important operations for our present consideration are the fundamental ones of Addition, Subtraction, Multiplication and Division. 38. Addition of positive and negative numbers. We have already seen in § 34 that addition of two negative numbers is performed in the same way as that of positive numbers. Just as + 3) + (+ 2) = + 5 - 3) + (- 2) = - 5. so Such operations can be confirmed by use of Fig. 9. The addition of a positive and a negative number can also be seen from Fig. 9. For example, (— 4) + (+ 3) is represented by starting at D, which represents — 4, and moving + 3 to the right to E, the result being — 1. Similarly (+ 3) + (— 7) as found by starting at F, J* TEACH YOURSELF ALGEBRA marking + 3, and since — 7 is a negative number, we move 7 divisions to the left to D to find the sum, which is (— 4). When the negative numbers involve letters, the procedure is the same. Thus: — 5a) + (+ a) = — 4« + 26) + (- 56) = - 36 — 2x) + {+ 6x) = 4*. 39. Subtraction. This operation presents a little more difficulty, since it is not easy at first to understand what is implied by the subtraction of a negative number, as, for example, ( + 6) - (- 2) or (- 2) - (- 5). This can be deduced from Fig. 9, but we will first obtain the rule by applying a fundamental law of addition and subtraction. Since then and 9 = 7+2 9-2 = 7 9-7 = 2, or, in general terms: if a = 6 -f c then a — b = c and a — c = b. We have seen above that (- 5) + (+ 3) = - 2 (- 2) = (- 5) + (4- 3). .*. from the above (- 2) - (- 5) = + 3 but we know that (— 2) + (4- 5) = + 3. ,\ comparing the two statements _ (_ 5) = (+ 5). A similar result will clearly hold whatever the numbers. .". We conclude that for any number a Similarly - (-«) = + «■ - (- la) = 4- la. POSITIVE AND NEGATIVE NUMBERS 53 Examples. 5x — (— 3*) = 5* + 3x = 8* - 26 - (— 46) = — 26 + 46 = 26. Graphical illustrations. To find (— 2) - (— 5). The rule can be deduced from the graphical representation of the number scale in Fig. 9 as follows: If we add a negative number we move to the left along the scale. Thus (- 2) + (- 5) = - 7. Consequently if we subtract a negative number we must move to the right. Starting from (— 2) and moving 5 to the right we reach + 3, i.e.. (- 2) - (- 5) = (+ 3). To find (-2) - (+5). When adding a positive number we move to the right. .*. when subtracting a positive number we move to the left. .*. starting from (—2) we move 5 divisions to the left and read (— 7). Summarising the rules for addition and subtraction we have: + (4- a) = (+ a 4 \-a) = (-a - (4- a) = (- - (- o) = (4- a . The student should compare these with rules for signs given in § 26. Exercise 8. 1. A lift starting from the ground-floor rises to the fourth floor. Then it descends to the second floor, rises to the sixth floor and finally descends to the ground-floor. Express its movements by using positive and negative numbers. 2. The movement of the mercury in a thermometer was as follows. Starting at 4- 8° C. it rose 2°, fell 14°, then rose 4° and finally fell 6°. Express these, using positive and negative signs, and find the final temperature. J4 TEACH YOURSELF ALGEBRA 3. How much higher than a temperature of — 15° C. is : (1) A temperature of — 4° C? (2) Freezing point? (3) + 15° C. ? 4. Using the number scale shown in Fig. 9 find : (1 By how much — 2 is greater than — 7 ? (2 By how much — 6 is less than — 1 ? (3 By how much + 3 is greater than — 5? 5. (1) What must be added to (— 3) to give (a) — 1, (2) What must be taken from (— 3) to give (— 8) ? 6. Write down the values of: + + -3 -2). -2). L 4)- ■I- — 2 2 -4 + 4 7. Simplify the following: + 2a - (- 5a). I _ 4* - (+ Zx). 3 + 3a6 — (— lab). 4 2x — 3y — 5y - 3*. (3a - 26) - (2« + 56). (3x - y) - (ix - 2y). 3* — (3y - ix). 8 (5 + x) - (6 - 2x) - (3* + 7). 8. (2) Subtract [x - 2y) from (3* — 4y). (2) Subtract (x — y + 2z) from (3* - 2y — 5z). 9. Fill in the brackets in the following: 3a - ( ) = 8a. 5*-( ) = -*• _ 3„ _ ( ) = 7a. 10. Write down the values of: (+ a). (2) - (- a). '31 + POSITIVE AND NEGATIVE NUMBERS 55 40. Multiplication. (1) Multiplication of (— a) by (+ b) and (+ a) by (- b). Consider, as a special case (— 2) x (+ 3). Since multiplication is a shortened form of addition, the meaning of (- 2) X (+ 3) is (— 2) + (— 2) + (- 2). This by §59= (-6). (-2) X (+3) = (-6). This can be applied to any pair of numbers, and so we may conclude that in general (- a) X (+ 6) = (- ab). Since the multiplication of two numbers can be taken in any order (§ 17) (+ a) X (- 6) = (- 6) x (+ a) and this by the above result is (— a6). (2) Multiplication of (— a) by (— b). Since (- a) X (+ b) = (- a6) and a negative number operates in the opposite sense to a positive number, it follows that (- a) X (- 6) = (+ a6). 41. Division. (1) Division of (+ a) by (+ b). Since (+ 4) x (+ 3) = + 12. (+ 12) - (+ 4) = (+ 3). Similarly + (a) -4- (+ 6) = + (|). (2) Division of (— o) by (+ b). Since (+ 4) X (- 3) = (- 12). (- 12) -r (+ 4) = (- 3). Similarly (- a) -4- (+ 6) = (— ^). (3) Division of (+ a) by Again (— 4) X and (+ 12) -^ Similarly -b). -*■ 3) = (+ 12) - 4) = (- 3). (+0-5- (-«-(-!) TEACH YOURSELF ALGEBRA -b. -3 (4) Division of (— a) by As above (+ 4) X (- 12) -f - (- 12). and in general - 3 = (+ 4). (- «)* (-&)=(+ j). 42. Summary of rules of signs for multiplication and division. Multiplication. + o — a — a + b) = -f ab -b) = -ab + b) = —ab - b) = + ab. Division. (+0)-r (+*>) = (+?) <+■> -(-»)" ("J) (_ fl )4-(+6)-(-f) (- ),-(_ b ) = (+°) These results can be summarised in the following mle : In the multiplication and division of positive and negative numbers, if the two numbers have the same sign the result is a positive number. If the signs are different, the result is a negative number. Or to remember these more readily the following slogan can be used: Like signs give + Unlike signs give — 43. Powers. Squares and square roots. When we square a number we multiply two numbers with the same signs. In accordance with the above rules, the product must be positive. POSITIVE AND NEGATIVE NUMBERS 57 Thus (+ a) X (+ a) = + a» (- a) X (- a) = + a": Consequently the square of any number is positive. It follows that when this operation is reversed and the square root of a' is required, this may be either (+ a) or (-a). To indicate this we use the sign ±, meaning " plus or minus ", i.e., Va* = ± a. Again (— a) 3 = (- a) X (- a) X (- a) = - a 9 and (_ a )* = (— a) X (— a) X (- a) X (— a) = + a*. From these and similar examples we may deduce that: An odd power of a negative number is negative. An even power of a negative number is positive. Exercise 9. 1. Write down the answers to the following: (2) (+ 12 X (— 3). (4) - 12 x - 3). (6 (+ 12 -f- (- 3). (8) (— 12 -^ (- 3). 2. Write down the answers to the following: 3. Write down the answers to the following: (1) (+ 12) x (+ 3). 3) (- 12) X + 3). (5 + 12) 4- (+ 3). (7 (- 12) + {+ 3). (1) (- 2a) X (+ 26). (2 (3) (+ 10*) X (- 2y). (4 (5) (+ 10*) -^ (- 2y). (6 4. Find the values of the following: (1) (- 4) X (+ 3) X (- 2). (2) (- a) X (+ 3a) X (- 2a). (3) I- I8xy) H- (- 8*}. (4) (— 24a«6*) -r (+ -tab). (- 2a) X (- 26). (- 10*) x (+ 2y). (- 10*) -r (- 2y). 58 TEACH YOURSELF ALGEBRA 6. Find the values of the following : (1) (- 5*) x (- 2x) x (- x). (2 a(a - b) - b(b - a). (3 -{a (-26) x (-6)}. (4) — a(a — 2b — c). 6. Find the simplified form of the following: (1) (+ 2a) x (- 56) x (- 26). (2 (- 4*)» - 2*(5* - 4). (3) x(y -z)- z(x -y)- y(x - z). 7. Write down the second, third, fourth and fifth powers of: (1) (-«). (2) (-2*). (3) (-!). 8. Write down the square roots of 81 and Qx* and the cube roots of — x 3 and — 8a". 9. Find the answers to the following: - 8x) x (- 2). - lOx) + (- 2). - 2xy) + (- x). + 06) - (- 3). + «<) + (- 4/). - 4**) -i- (— 2**). - 2x*) X \- 4x). + 15x^y) -r (- o.vv). 9 — 12fl*6 s ) 4- (+ 3a6). (10 - 24a 3 6c«) + (- 4a6c). 10. Write down the values of: (1) {(- a) x (- 6)} -5- (- a). (2) (+ £) X (- *)«. (3) (- 6*)* X (- x) 3 4- (- 2x)«. CHAPTER V SIMPLE EQUATIONS 44. Meaning of an equation. If it is known that 5 times a certain number is 40, a simple process in Arithemtic enables us to calculate that the number is 8. In algebraic form the problem could be expressed as follows : Let n = the unknown number. Then the question can be put in this way : If 5« = 40, what is the value of m? The statement 5m = 40 is called an equation. It is a statement of equality, but it also implies that a value of n is required which will make the left-hand side of the equation equal to the right, or as we say " satisfies the equation ". Trie process of finding the value of n which thus satisfies the equation is called " solving the equation ". The solution of the above equation involves no more than the division of the right-hand side by the coefficient of m, and could be stated thus: 5m = 40. n = 40 -i- 5. n = 8. The solution of an equation is rarely so simple as this. Equations usually consist of more or less complicated ex- pressions on both sides of the equation. By various opera- tions we aim at reducing the equation to the simple lorm above. The value of the unknown letter is then easily found. These operations will be illustrated in the examples which follow. 45. Solving an equation. Example I. 1/8 limes a number is decreased by 5 the result is 123. What is the number? 59 6o TEACH YOURSELF ALGEBRA This simple problem could be solved mentally, but it will serve as an introduction to the process of solution. Let n = the number. Then 8m — 5 is the expression which states algebraically " 8 times the number decreased by 5 ". But this is equal to 123. Hence we can form the equation 8m - 5 = 123. This is the first step that must always be taken — to formu- late the equation. Then we proceed to the solution — i.e., to find the value of n which satisfies it. Now, the above statement means that 123 is 5 less than 8 times the number, or, if 123 be increased by 5, it is equal to 8 times the number. .'. we can write the equation in this form 8n = 123 + 5. Thus we have practically reached the form we wanted, after which we can readily find the solution. This step was reached in effect, by transferring the 5 to the right-hand side, leaving only a multiple of n, the unknown number, on the left side. In this transference the argument involved changing the sign of the 5. The same result could be obtained as follows: Since 8m — 5 = 123, if each side be Increased by 5 the result will be that we shall be left with 8» only on the left-hand side and the two sides will still be equal. We shall have as our equation: 8m -5 + 5 = 123 + 5 whence 8» = 128 and n= 16. This device is employed in the solution of practically every equation and it depends on the fact that: (A) If the same number be added to, or subtracted from, both sides of an equation, the two sides will again be equal. As a working rule this is equivalent to transferring a number from one side of an equation to the other at the same time changing its sign — i.e., change -f- to — and — to -f . SIMPLE EQUATIONS 6i A principle similar to the above which will be employed later, is: (B) If both sides of an equation are multiplied or divided by the same number, the two sides of the new equation will be equal. If the multiplier is — I, both sides change signs. Example 2. In §4 we saw that three consecutive odd numbers could be expressed algebraically by 2m + 1, 2m + 3, 2n + 5 where n is any integer. Now suppose we wish to solve this problem: The sum of three consecutive odd ntimbers is 81. What are the numbers ? As stated above, the first step is to form an equation. This usually means putting into algebraic form the facts which are given about the unknown number or numbers. We first, as above, represent the three odd numbers by 2m + 1, 2m + 3, 2m + 5. Then we express algebraically, the fact that their sum is 81. Hence we get the equation : (2m + 1) + (2m -(- 3) + (2m + 5) = 81. The use of the brackets helps to make the statement clear. We now remove the brackets and get: 2m + 1 + 2m -f 3 + 2m + 5 = 81. Adding like terms G» + 9.= 81 whence, as above (in = 81—9 or 6m = 72. n = 12. We now obtain the odd numbers by substitution of m = 12 in 2m + 1, 2m + 3, 2m + 5. .'. the numbers are 25, 27. 29. This should be checked by ascertaining that their sum is 81. 46. Worked examples. Equations arise out of practical problems in a variety of ways, and examples will be given later, but it is desirable that the student should first have sufficient practice in the (n TEACH YOURSELF ALGEBRA methods of solving equations. Examples of equations will therefore be worked out and provided for practice which will have no relation to any special problems. It is usual in such practice equations to use letters at the end of the alphabet, x,y and z, to represent the unknown numbers, and, when necessary, letters at the beginning of the alphabet, a, b, c, etc., to represent known numbers. This choice of letters is due to Descartes (seventeenth century). Example I. Solve the equation: 6* — 5 = 2x + 9. The general plan adopted is to collect tlie terms involving the unknown number, x, on Hie left side, and the oilier terms on the riglU. Transferring the x term from the right side we get: 6* — 2x — 5 = 9. Transferring the — 5, 6x - 2x = 9 + 5 4x = 14 and X = r N ote . — With practice the two transference steps could be taken together. Check. The accuracy of the solution to an equation can always be checked by substituting the value found in both sides of the original equation. In the above case: Left side (o X ?) - 5 = 16. Right side (2 X ?) + 9 = 16. The two sides are equal. x = ~ satisfies the equation. Example 2. Solve the equation : 10(* - 4) = 4(2* - 1) + 5. SIMPLE EQUATIONS 63 First simplify both sides by removing brackets. Then 10* — 40 = 8x — 4 + 5. Transferring 8* to the left side and — 40 to the right, 10* — 8* = 40 — 4 + 5. 2x = 41 and x = 201. Check. Left side 10(201 - 4) = 10 X 161 = 165. Right side 4(2 x 20J - 1) + 5 = 160 I 5 = 165. x = 20£ satisfies the equation. Example 3. Solve the equation: 3* x ox . T + 2 = T~ 3 - When the equation involves fractions, the first step, in general, towards simplification is to " clear the fractions ". This is effected by multiplying throughout by such a number that the fractions disappear. This is justified by Principle B, § 45. The smallest number which will thus clear the fractions is the L.C.M. of their denominators, in this case 20. Multiplying every term on both sides by 20 we get: (I X 20) + (I X 20) = ( *j X 20) - (3 X 20). 12* + 10* = 25* — 60. 12* + 10* — 25* = — 60 - 3* = - 60. * = — 60 -j- — 3. x = 20 (by the law of signs). This should be checked as in previous examples. Example 4. Solve the equation : Multiplying throughout by 12 48* - 4(* - 2) = 60 + 3(2* + 1). (§ 28.) 64 TEACH YOURSELF ALGEBRA Clearing brackets 48* — 4x + 8 = 60 -f 6x + 3. 48* — 4x - 6* = 60 + 3 - 8. 38* = 55 55 38' Check by substitution. and x = Exercise 10. Solve the following equations: 1. 2. 3. 4. 5. (a) 14* = 35. fx = 24. 3x = - 48. - \x = - 40. *-* 6. (« 7. u 8. u 9. {" 10. a (b 11. 5x + 8 = 24. 3* - 4-7 = 2-8. 3x - 5 = 2x + 3. 2* + 6 = 14 - 3x. 3y - 1-5 = ly - 8-7. 4-8* + 52 = 3-2x — 20. (a) 3y - ^ - 2y + 6. (6) (b l-5x = 30. (b fx = 50. (6 -2* = 20. \b -Jx = 24. «« — <*. lOx - 9 = 41. 2-5* -f 50 = 80. (b) 6y + 11 = 3y + 15. (b) z + 20 = 52 - 44. x + ^ = 20. 12. (a) o 3 (b) 0-1 + 0-2 = 5. 13. 4(2x - 5) = 3(2x + 8). 14. 15. 16. 17. 18. 19. 3* - 2(x + 4) = 5* - 28. = 20. = 2y + 12. = 4(x + 1) - (x - 5). 20. (a x + 5) — 3(x - 6 5(v- l)-2fcy+6 ~x- l) + 3(x + 4 x-7)-3(2x-4) = 4(x + 3). 3y + 6) - *(2y - 4) = 20. l-j = 0. {b) |-3 = 8. 21. (a) ^-(x-2) = 12. W 2 3 6 ' 22. 2x— 5 3*-l •1 3 2* 23. SIMPLE EQUATIONS 7x — 6 5x + 3 6x — 1 _ .J. 63 4 7 14 24. i(x + 3) - f (x - 3) = * + 9. 25. 12(5 — x) - 3(3x - 4) = 23. 26. f(x - 1) - J(5 - x) = 2) - 3(x - 3). ..{-if-*-* 28. 2x + 4 3x — 5 = 1. 1-5 2-5 29. Solve for » the equation 2» = 0-58(12 — n). 30. For what value of r is 18-4 equal to 2(3-5r — 1) ? 31. Find x when — = jj. x 2 32. Find c if ^ = 3-8. 2c 33. If C = ? find F when C = 8, i? = 4-5. 34. For what value of x is 3 (x - 5) equal to ** £ 3 ? 47. Problems leading to simple equations. The methods of solving problems by means of simple equations are illustrated by the following examples. The general method of procedure is: (1) Having decided which is the unknown quantity, represent it by a letter, such as x, stating clearly the units employed when necessary. (2) Form an equation which represents the facts provided by the problem about the unknown quantity. (3) Solve the equation. Example I. At a dance party there were 10 more women than men. The men paid 20/> each, the women \5p each, and the total receipts were £16-20. How many men and women were there at the party? There are two unknown quantities: the number of men and the number of women. But if the number of men is known, the number of women is 10 more. 66 TEACH YOURSELF ALGEBRA .*. Let * — the number of men. Then * + 10 = the number of women. The facts supplied are represented as follows: 20* mm the money paid by the men, in pence; 15(* + 10) = the money paid by the women, in pence; 1620 = the total amount paid, in pence. The equation which connects these is consequently: 20* + 15(* -f 10) = 1620. 20* + 15* + 150 = 1620 35* = 1620 - 150 35* = 1470. x = 42. ,". the number of women is x + 10 = 52. .'. the solution is 42 men, 52 women. Example 2. A motorist travels from a town A to another town B at an average speed of 24 miles per hour. On his return journey his average speed is 32 miles per hour. He takes 7 hours for the double journey, exclusive of stops. How far is it from A to B in miles ? The unknown quantity is distance from A to B. Let * = distance in miles. Now distance = speed x time. distance j— = time. speed Then time for 1st journey = 55. „ 2nd „ But total time is 7 hours. .". the equation is * 32" -i + - = 7 24 + 32 '• To clear of fractions multiply throughout by 96. •"• (& X96 ) + (31 X96 ) = 7><96 and 4* + 3* = 7 X 96. 7* = 7 X 96 x = 96 miles. SIMPLE EQUATIONS 67 This should be checked by applying the conditions in the question. Example 3. The weekly wages of a man and woman engaged on the same kind of work were £4-80 and £3-70 respectively. It was agreed to increase the two wages by the same amount so that the man's wage was J of tlie woman's wage. What increase was given ? The unknown quantity is the money to be added to the wages. Let * = the amount of increase in pence. Then (480 + *) pence = the man's new wages. (370 -j- *) „ = the woman's new wages. Then by the data 480 + * = 2(370 + *). Clearing fractions 5(480 + *) = 6(370 + *). 2400 + 5* = 2220 + 6*. 2400 — 2220 = 6* — 5x and 180 = x. ,*. the weekly increase is 180p or £1-80. This should be checked by adding it to each of the weekly wages and ascertaining if one is f of the other. Exercise II. 1. From three times a certain number, n, 6 is subtracted. The result is equal to twice the number together with 6. What is the value of n ? 2. There is a number such that when it is multiplied by 5 and then 14 is subtracted, the result is 348-5. Find the number. 3. From 5 times a certain number 189 is subtracted, and the remainder is one half the original number. What is that number? 4. One-third of a number added to four-fifths of itself is equal to 17. What is the number? 5. When 9 is subtracted from 6 times a certain number, the result is 45 more than twice the number. Find the number. 68 TEACH YOURSELF ALGEBRA 6. The sum of three consecutive odd numbers is 69. What are the numbers ? 7. A man walks from one town to another at an average speed of 2J miles an hour. On the return he quickens his average speed to 3 miles an hour. The time taken for the double journey was 7 hrs. 20 min. How far are the two towns apart ? 8. The sum of a number and 4 per cent, of itself is 41-6. What is the number? 9. The perimeter of a rectangle is* 44 ins. If one of the two adjacent sides is 1-8 ins. longer than the other, what are the lengths of the sides ? 10. Some men agree to pay equally for the use of a boat, and each pays 15p. If there had been two more men in the party, each would have paid lOp. How many men were there and how much was the liire of the boat? 11. A man distribute £2 among 20 boys, giving 5p each to some and 25p to the rest. How many boys received 25p each? IS. A man is four times as old as his son. In four years time he will be three times as old. What are their ages now? 13. The connection between the degrees on theCentigrade and Fahrenheit thermometers is that x° C. = [-jr + 32 J °F. What number of degrees Centigrade is equivalent to 86° F.? . . 14. A bookseller buys 120 volumes of a certain series of books. He sells some at the published price of 18p each and disposes of the remainder during a sale at 12p each. If his total receipts are £19-20, find how many volumes were sold at each price. 15. A bus is carrying 32 passengers, some with 3p tickets and the remainder with 5p tickets. If the total receipts from these passengers are £1-14, find the number of 3p fares. CHAPTER VI FORMULAE 48. Practical Importance of formulae. One of the most important applications of elementary Algebra is to the use of formulae. In every form of applied science and mathematics, such as mechanical engineering, electrical engineering, aeroplane construction, etc., formulae are constantly employed, and their interpretation and manipulation are essential. 49. Treatment of formulae. In the first chapter of this book some very easy examples of formulae were introduced. With the assistance of a greater knowledge of algebraical symbols and operations and with increasing skill in their use, the student can now proceed to more difficult types. Formulae involve three operations: (1) Construction; (2) manipulation; (3) evaluation. The construction of formulae cannot be indicated by any specified rules or methods. A knowledge of the prin- ciples of Algebra and skill in their application are necessary. But in general the student is concerned with formulae which have already been evolved. What he needs is skill in using them; and as his knowledge of Algebra increases, so he becomes better equipped for dealing with new examples. The manipulation and evaluation of formulae are closely associated. A formula may need to be re-arranged or simplified before any substitution of values may be made. Experience alone will guide the student as to what manipula- tion is desirable in order to reach a form which is the most suitable for evaluation or some other purpose. Clear arrangement of working is always essential for accuracy. G (ALG.) 69 7° TEACH YOURSELF ALGEBRA 50. Worked examples. Example I. Find a formula for the total area (A) of the surface of a sqtiare pyramid as in Fig. 10 when AB = a ins. and OQ = d ins. OQ is perpendicular to AB, and represents the height of the &AOB. The total area is made up of: 1) Area of base. [2) Areas of the four As, of which AOB is one. Area of base = a* sq. in. Area of each A = \ad. .-. area of all As = 4 X \ad — 2ad. .', total area of pyramid = «* + 2ad or A =o(o+ 2d). Example 2. If L = ™^~ ° - l - find L when t = 8-5, w = 115. W = 380, T - 28-5. Substituting in Fig. 10. w t. _ 380(28-5 - 8-5) __ _ . L - 115 380 X 20 - _ " -TTT- - 8-5 = 06 — 8-5 (approx.) = 57-5 (approx.). Exercise 12. 1. If s = ut + J// 2 , find s when « = 15, t = 5, / = 8. 2. The volume of a cone, V, is given by the formula V = btr*h, where r = radius of base, /» = height of cone. 22 Find V when r = 3-5, h = 12, it = y. 3. The volume of a sphere is given by the formula 4 22 V = 5itr», where r = radius. Find V when r = 3, it = -=-. FORMULAE 7' Wv* 4. If £ = -£-, find E when W = 15-5, v = 18-8. and £ = 32. F A- t 6. In electrical engineering the formula C = „ "T is used. Find C when E = 17-6, e = 1-5, i? = 28-4, r = 2-6. 6. In a suspension bridge the length of the cable employed is given by the formula L = I + -=p where L = length of the cable, d = dip of the centre of the cable, / = length of the span of the bridge, all measurements being in yards. Find L when d = 6, /=56. 4/ibd* 7. A formula for the loading of beams is W = .— . Find W when k = 45, b = 2, d = J. / = 20. 8 . If 8 = »(» + i)P» + 1) find s when w = 8 . 9. The following formula is used in connection with pile Wh driving, L = ,,,,, , „ .. Find L when W = 5, d = 1-5, aflr + /*) P = 19, A = 4-5. 10. If R = W(* + 3/), find /? when W - 210, * = 6-5. < = 0-04. 11. The horse power, H, of a steam engine is given by the formula H = UJgg. Find ff when p = 18, / = 2, A = 80, 2V = 360. 51. Transformation of formulae. In the formulae which have been examined it will be seen that one quantity is expressed In terms of other quantities and the formula expresses the relations between them. Thus in the formula for the volume of a cone <Y = Jit r 2 h, Exercise 12, No. 2) this volume is expressed in terms of the height and the radius of the base. But it may be necessary to express the height of the cone 7a TEACH YOURSELF ALGEBRA in terms of the volume and the radius ot the base. In that case we would write the formula in the form: that is, the formula has been transformed. When one quantity is expressed in terms of others, as in V = JwV», the quantity thus expressed, in this case V, is sometimes called the subject of the formula. When the formula was transformed into h = 3V ■nr v the subject of the formula is now h. This process of trans- formation has been termed by Prof. Sir Percy Nunn " changing the subject of the formula ". The transformation of formulae often requires skill and experience in algebraical manipulation; the following examples will help to illustrate the methods to he followed. 52. Worked examples. Example I. From the formula T = —ji find: 1) / in terms of the other quantities. r«5g! 1 ~ 16* From clearing the fraction 16T = nfd 3 . . or nfd 3 = 167\ dividing throughout by mi 8 From (1) dividing throughout by T.f 16T w (1) d: FORMULAE 73 Example 2. Transform the formula into one which expresses d in terms of the other quantities. 8d* L=l + W or Clearing fractions: 31L = 3/* + 8d* 8d 2 = 31L - 31*. 31L - 3/» d l = d s w 3/L - 3/* 8 Example 3. The velocity, V, of water flowing through a pipe, occurs in (lie formula, I V 7 Change (lie subject of the formula to V. Writing the formula as / V* 0-03 B x- 2 - = h. Divide both sides by 003 ^, V* , . 0-03L 2i = H ^-D- V* = Ax hD 0-D3/.' 2ghD 0-03L' D 0031 v= l2gW V003L" 74 TEACH YOURSELF ALGEBRA Example 4. If a — b = x(c — nd) find n in terms of the other letters. Attention should be fixed on the term containing n, viz. nd. This we try to isolate from the other terms. .*. divide both sides by x. a — b Then Transferring c a-b ■ = c — nd. x — c= — nd. , a-b nd = c . I / a-b\ n = -A c --ir)- Example 5. The lime of vibration of a simple pendulum is given by the formula t: Find I in terms of the other quantities. •-•JL Square both sides. Then <* = 47t»xi. or t*g = 4tc»/ 4**/ - gt*. Exercise 13. 1. The formula for the area (A) of a circle, in terms of its radius (r) is A = rer*. Change the subject of the formula to express r in terms of the area. 2. Transform the formula for the volume of a sphere- viz. V = Jitr 3 (see Exercise 12, No. 3)— into a formula in which r is expressed in terms of the volume. FORMULAE 7S 3. Change the formula for the volume of a cone — viz., V = \-ktVi — to a formula in which the subject is r. 4. The horse power of a motor is given by the formula H = gfjg. Express this as a formula for C. 5. The lifting force of an electro-magnet is given by the B % A formula F = 5 , where F is the force. Transform this into a formula of which the subject is B. 6. The amount of sag, d, in a beam under certain con- . . WP ditions is given by the formula d = r^, Express this as a formula expressing / in terms of the other quantities. 7. If ti* = m* + 2/s, express s in terms of u, v, and /. Find the value of s when « = 15, v = 20, and / = 5. 8. There is an electrical formula / = =. Express this (1) as a formula for V and (2) as a formula for R. Find / if V = 2 and R = 20. 9. If nV + 1 = AT?, rearrange the expression so that it becomes a formula for n. Find the value of n when iV = 25, R = 2, r = 0-81. 10. The relation of the volume (t/) of a mass of gas to the pressure (p) on it is given by the law pv = k. In a certain experiment when p = 84, v = 12. Find the value of k and then express the formula giving v in terms of p and the value of k. 53. Literal equations. The operations employed in changing the subject of a formula are the same in principle as those used in the solution of equations. One essential difference from the equations dealt with in Chapter V is that whereas these were concerned with obtaining numerical values when solving the equations in the formula the quantity which is the subject of the formula is expressed in terms of other quantities, and its numerical value is not determined, except when the numerical values of these quantities are known. 76 TEACH YOURSELF ALGEBRA It is frequently necessary, however, to solve equations in which the values of the unknown quantities will be found in terms of letters which occur in the equation. Such equations are termed literal equations. The methods of solution are the same in principle as those employed in Chapter V. They are illustrated in the following examples. 54. Worked examples. Example I. Solve the equation 5x — a = 2x — b. As pointed out previously (§ 46), x is understood as stand- ing for the unknown quantity and the use of the letters a and b marks the difference between this kind of equation and those of Chapter V. The methods by which the equation is solved are the same, however, a and b being treated in the same way as ordinary numerals. In 5x — a = 2x — b transferring 2x and — a respectively from one side to the other, changing the signs in so doing, 5x — 2x = a — b. 3.r = a — b a-b and x = — . — . Example 2. Solve for x a(x - 2) = 5* - (a -f b). Removing brackets ax — 2a = 5x — a — b. Transferring ax — 5x = 2a — a — b or ax — 5x = a — b. This introduces a point of difference from numerical equations. With the latter we add the terms involving x by adding their coefficients. The addition in this case cannot, however, be made arithmetically. Algebraically the sum of the coefficients of x is a + (— 6) or a — 5. .", we write (a — 5)x = a — b. Dividing both sides by the coefficient of x a-b x = o-5' FORMULAE Exercise 14. Solve the following equations for *: 1. 5x - 4a = 0. 2. 5x — 3a = 7a. 3. 8x — p = 3* + 4p. 4. 3* + 26 = 2(x + 36). 5. ax -f- 6 = 3a — 6. 6. b[x -p)=c. 7. 2a — b = b — bx. 8. 3(ax - 2) + 266 = 66. 9. p(x -q)=x{p- q). 77 10. ax 2 " -4 = ■J-* 11. ax - -46 = bx - 6. 12. \+ X 3a~ 4—x. 18. a(x -a) = b(x + 6). X 1 3* N. m H inn CHAPTER VII SIMULTANEOUS EQUATIONS 55. Simple equations with two unknown quantities. The simple equations considered in Chapter V contained only one unknown quantity whose value it was required to determine. But many of the formulae quoted in the pre- vious chapter contain several quantities. Cases may therefore occur in which it will be required to find the value of more than one of these. Similarly problems arise which, for their solution, involve the determination of more than one unknown. 56. Solution of simultaneous equations. A simple problem will serve to illustrate the above state- ment. Suppose we are told The sum of two numbers is 10 ; what are the numbers ? Let the two numbers be represented by x and y. Then we know that x + y = 10. It is evident that there is an infinite number of solutions of this equation, such as (1, 9), (2. 8), (3-5. G-5), etc. If the equation be written in the form: x= 10 — v. This gives x In terms of y. Whatever value be given to y In this equation, a corre- sponding value of x can be found, and each pair of values furnishes a solution of the equation. If a second condition has to be fulfilled we can determine which of these pairs satisfies it. For example, if, in addition to the statement that the sum of the numbers is 10, we are also told that one of them 78 SIMULTANEOUS EQUATIONS 79 is four times the other; then there is only one set of the pairs of values referred to above which will satisfy both the conditions. For if and x+y=10 x = 4y (1) (2) substituting for x in equation (1) we get: 4y + y = 10 5y= 10 y = 2. and since then * = 4y, x = 8. .". the solution which satisfies both equations simultaneously is x = 8,y = 2, and clearly there is no other solution. For this reason such equations are called simultaneous equations. It is evident that if there are two unknown quantities whose values are required, tt must be possible to form two separate equations connecting them. The methods employed in solving these equations are shown in the following examples. 57. Worked examples. Example I. Solve the equations : 2x + y = 21 3* + 4y = 44 (1) (2) In the method employed in this example we begin by obtaining one letter in terms of the other. The more convenient one is chosen, and in this case from equation (1) 2x + y = 21, we get: y = 21 - 2x (3) We could have found x in terms of y, but this would involve fractions and is not so convenient. Substituting in equation (2) the value of y thus obtained from equation (1) 3* + 4(21 - 2x) = 44. So TEACH YOURSELF ALGEBRA Thus we reach a simple equation with one unknown. This is solved as previously: 3* + 84 — 8* = 44 3* — 8* = 44 — 84 — 5* = — 40 or 5x = 40 and * = 8. Substituting this value of * in equation (3) we can find y. Thus y = 21 - (2 x 8). y = 5. /. the solution is x = 8. y = 5. These values should be checked by substitution in both of the given equations. Example 2. In the following example a second method Is shown which can frequently be employed to advantage. Solve the equations : x + y = i5 .... a; 3*-y = 21 .... (2 It wiU be seen that if the left sides of the two equations were added, y would be eliminated, since we would get (+ y) + ( — y) = 0. Thus only * would remain. It is clear also that the sum of the two left sides of the equations must equal the sum of the two right sides, i.e., (x+y) + (3* -;y) = 15-t-21, whence x + Zx = 36 4x = 36. * = 9. Substituting this value of x in equation (2) or, if easier, In equation (1) (3 X 9) — y=>21. 27— y = 2\ -y=-6. y = 6. .'. the solution is x = 9, y = 6. Example 3. In the following example both of the above methods are employed. Solve the equations : 2* + 3? = 42 . . . . (1) 5x—y = 20 .... (2) SIMULTANEOUS EQUATIONS 8i 1st method. Substitution. From (2) -y = 20-5*. y = 5x — 20 . . . . (3) Substituting in (1) : 2x + 3(5* - 20) = 42 2x + 15* - 60 = 42. 17* = 102 and * = 6. Substituting for * in (3) : y = (5 x 6) - 20 and y = 10. ,*. the solution is x = 6, y = 10. 2nd method. Elimination. In this example neither letter can be eliminated by addi- tion of the left sides of the equation, as in example 2. But by multiplying both sides of equation (2) by 3, y can be eliminated. We proceed thus : 2* + 3y = 42 .... (1) 5x-y = 2Q .... (2) Multiplying by 3 throughout in (2) 15* — 3y = 60 . . . . (3) Adding (1) and (3) (2* -f Zy) + (15* - Zy) = 42 + 60 17* = 102 and x = 6. From this y can be found as before. Note. — * could have been eliminated from (1) and (2) as follows : (a) Multiply throughout in (1) by 5. (b) .. „ (2) by 2. Then we get: 10* + 15y = 210 10* — 2y = 40. Subtracting (10* + 15y) - (10* - 2y) = 210 - 40. 10* + 15y - 10* 4- 2y = 170 whence 17y = 170 and y = |0. ~ 82 TEACH YOURSELF ALGEBRA Comparison of the methods. Of these two methods, that of substitution is the sounder and more general. In practice equations are seldom easily dealt with by the elimination method. Example 4. Find values 0/ R t and R t which will satisfy the equations : 0-5/?, + l-2R t = 1-486 . . 4-5/?i - 2R t = 4-67 . . From (1) 0-5/?, = 1-486 — l-2R t . _ 1-486 -l-2Ri "1 - ' (1) (2) Substituting in (2) : 0-5 = 2-972 — 2-4/?,. 4-5(2-972 - 2-47?,) — 2R t = 4-67 13-374 - 10-8K, - 2R t = 4-67 - 12-8R, = - 8-704 R - — K * 12-8 /?, = 0-68. Substituting in (2) : 4-5K, - (2 X 0-68) = 4-67 4-57?! - 1-30 = 4-67 4-57? 1 = 603. "1 — 4.5 = 1-34. /. the solution is R, = 1-34. R t = 0-68. Exercise 15. Solve the following equations : 1. y = 2x. 2.y = 3x- 7. 3* + 2y = 21. 3. x = 5y — 3. 3* _ Sy = 12. 5. 3* - 2y = 7. x + 2y = 5. 5x — 3y = 1. 4. x — 3; = 5. 4*— ;y = 2x+ 13. 6. 2x-y = 3. x + 2y= 14. SIMULTANEOUS EQUATIONS 83 7. 2* — y = 10. 3* + 2y = 29. 9. 3* — 2(y 4- 3) = 2. 2(x — 3) 4- 4 = 3y - 5. 13. |— 2a = 5. 8. 2(* - 4) = 3(y - 3). jy - 2x = - 13. 10. 1 - * = 2. 12. |+6=1. <L 6 31 3fl -3=2- 14. 4(1 -p)=7g+ 8p. 3a + 46 = 6. 6p + q 4- 8 = 0. 15. 7* + 3(v-3) = 5(*+.y). 7(*-lJ-6y=5(*-y). 16 g~3 = g- 17. 2(3«-6)=5(a-2). 26 =5 - 3(a 4- 46) = 2(6 - 3). 18. 0-1* + 0-2y = - 0-2. 19. 1-28* - 0-75> = 1. 1-8* — 0-4>> = 10-6. 0-25* 4- l-25y = 17. 20. 2-5* + 3Ty = 13-365. 21. *-=-! + :? — ^ = 3 - O it 8-2,-1.^ = 7-02. L^_^_zi = l 22. 2P — 5Q = 2. 3P 4- 100 = 8-6. 23. 3(x-y)-±(x+y) = 30. x+y+^(x-y)=22. 3^2 3 6 '■ 58. Problems leading to simultaneous equations. Many problems require for their solution the determina- tion of two unknown quantities. The general method of solution is similar to that employed when there is one unknown, but with the important difference that when &♦ TEACH YOURSELF ALGEBRA there are two unknowns to be determined, two equations must be formed from the data. The following examples illustrate the methods employed. 59. Worked examples. Example I. There are two numbers such that the sum of the first and three limes the second is 53, wk ilt the difference between A limes the first and twice tlie second is 2. Find the numbers. Let x = one number. „ y = the second. Then from the first set of facts * + 3y = 63 . . . . (1) From the second 4* — 2y = 2 . . . . (2) From (1) x = 53 - 3y. Substituting in (2) : 4(53 - 3y) — 2y = 2. 212 - 12y - 2y = 2. Collecting - Uy = - 210. -210 and y = 15. Substituting in * = 53 — 3y x = 53 - (3 X 15). * = 8. .". the numbers are 8 and 15. Example 2. In the equation y = mx + b it is known that the equation is satisfied by two pairs of values of x and y — viz. when x = 4, y = 6, and „ x = 2-4, y = 4-5. What are the values of m and b P This is an example of an important practical problem. It means that there is a law connecting x and y, the law involving m and b, which are constants. These constants must be determined before the law can be stated. They are therefore the unknown numbers in this case. The equations connecting them are obtained by substituting the given pairs of values of x and y. SIMULTANEOUS EQUATIONS 85 (1) When x = 4, y = 6. ,*. on substitution 6 = Am + b (1) (2) When x = 2-4, y = 4-5. 4-5 = 2-4w + b .... (2) These are to be solved simultaneously for a and b. It is clearly a case for using the method of elimination. Subtracting (2) from (1): 6 - 4-6 = {Am + b) - (2-Am + b) or 6 — 4-5 = Am + b — 2-Am — b. 1-5 = l-6w. 1-5 15 m Substituting in (1) 1-6 or T6- • a . and 15\ + 6 = 6. 6 = 6 15 4 .'. the solution is m= |6 , fa = 9 4 - Substituting these we get n ym. y = mx + b 15 , 9 T6* + 4 or clearing fractions I6y = I5x + 36. Example 3. A bookseller has a number of books the pub- lished price of which is 25p. After selling a certain number at this price lie sells the remainder at 20p each, and his total receipts were £11. If the numbers sold at the two prices were reversed, he would have received £11-50. How many books liad he in all and how many were originally sold at 25p? Let x = the number originally sold at 25p; .. y = ... ., ., 20p. The amounts received for these were 25* pence and 20y pence and their total value was £11 or 1100 pence. 25* + 20y = 1100. 86 TEACH YOURSELF ALGEBRA When the numbers are reversed lie receives 20* and 25y pence and their total value is now £1 1-50 or 1 150 pence. 20* -f- 25> = 1150 .". the equations to be solved simultaneously are: 25*+20y = 1100 ... (1) 20* + 25y = 1150 ... (2) From (1) 25* = 1100 - 20y , „ 1100 - 20y and * = 25 Substituting in (2): 20(1100 -*>y) +25y=im 4(220 - 4y) + 25y = 1 150 and 880 - 16y + 25y = 1 150 9y = 270 y = 30 110 — 20y 25 10 — 600 500 25 = 25 Since * = on substitution * = x = 20. .'. the total number of books sold was 30 -f 20 = 50, and the number originally sold at 25p was 20. Exercise 16. 1 . There are two numbers, * and y, such that the sum of 2* and y is 34, while the sum of * and 2y is 32. What are the numbers? 2. There are two numbers such that if to 3 times the first, twice the other is added, the sum is 72. Also if from 5 times the first number, 3 times the other is subtracted, the result is 44. What are the numbers ? 3. One number is greater by 6 than twice another number, but 3 times the smaller number exceeds the greater by one. Find the numbers. 4. If from twice the greater of two numbers 17 is sub- tracted, the result is half the other number. If from half SIMULTANEOUS EQUATIONS 87 the greater number 1 is subtracted, the result is two thirds of the smaller number. What are the numbers ? 5. In the equation y = mx + b, when * = 3, y — 3 and when x = 5, y = 7. Find the values of m and b and write down the equation. Then find y when * == 6. 6. Two quantities P and Q are connected by the formula: When Q = 5, P = 14 and when Q = 2. P = 20. Find m and b. 7. The force (E) applied to a machine and the resistance (R) to be overcome are connected by the law E = a + bR. It is found that when E = 3-5, R = 5 and when E = 5-3, R = 8. Find a and b. Then find E when R = 8. 8. It is known that y = ax 2 -f bx* ; when * = 2, y = 5-6, and when * = 3, y = 25. Find the values of a and b. 9. The perimeter of a rectangular lawn is 192 ft. It is reduced in size so that the length is four-fifths and the breadth is three-fourths of the original dimensions. The perimeter is then 150 ft. What were the original length and breadth ? 10. The bill for the telephone for a quarter can be expressed in the form C = " + 100 where C is the total cost in pounds, a is a fixed charge, n the number of calls and b the price of each call in pence. When the number of call is 104, the bill came to £5-83, and when the number was 67 the bill was £5-09. Find the fixed charge and the cost of each call. 11. The cost of 4 ties and 6 pairs of socks was £3-40, while that of 5 ties and 8 pairs of socks was £4-37. What were the prices of a tie and a pair of socks respectively? 12. The formula S = til -\- If? gives the distance (S) passed over by a moving body in t sees. In 4 sees, the body moves 88 ft. „ 6 „ „ 168 ft. Find the values of « and /and then find how far the body moves in 5 sees. CHAPTER VIII GRAPHICAL REPRESENTATION OF QUANTITIES 60. The object of graphical work. The graphical or pictorial representation of statistics and the comparison of magnitude by means of various graphical devices are familiar features of modem life. Rows and columns of figures or groups of large or very small numbers are not always readily grasped. Accordingly, the use of lines or columns or other figures drawn to scale which appeal to the brain through the eye, is found to be an effective method of enabling many people to realise, not only the quantities themselves but the deductions to be drawn from them. The graphical method, developed on scientific and mathe- matical lines, is also largely employed in mathematics and science to illustrate certain important underlying principles. The following examples, arranged progressively, will assist the student to understand the various forms of graphical representation. 61. The column graph. The first example, Fig. 11, is a reproduction of an actual graphical advertisement, is- sued by the Cement Makers' Federation to show the vari- ations in the deliveries of cement over a period of 11 years. It is an example of what is called a column graph. Along the straight line OX equal spaces are marked off which indicate, in succession, the eleven years from 1929 to 1939 inclusive. Fig. 11. In the spaces so formed are 88 Y MI OION TONS 1329 30 31 32 33 34 35 36 37 33 39 GRAPHICAL REPRESENTATION OF QUANTITIES 89 constructed a series of columns of the same width and whose height, on the scale chosen, represents the number of tons of cement manufactured in each year. In order that these heights can be easily read a scale is marked along the OY, perpendicular to OX, in which each unit represents a million tons. This assemblage of columns conveys very effectively, in a general way, the rise and fall of the production of cement during the eleven years. It also makes it possible to ascer- tain at a glance the approximate amount of cement pro- duced in any one of the years. 65 60 55 50 a .::::.:...: 1 . .: .: . -•'■■• '- 45 40 O L_ X J Jan Feb Mar Apl May Jne Jly Aug Sep Oct Nov Dee Fig. 12. 62. A straight-line graph. The next method illustrated is more commonly used than that of the column graph. The following table shows the mean temperature for each month throughout the year. 9° II AL i I UUH itL 1- A .ijt IRA Month : Jin. Feb. Mar. Apr. Miy. June. July. Aug. Sept. Oct. Nov. Deo. Mean Temp. 41-9 41-5 44-1 48-4 66-8 60-1 as-; G3-S til S'.'-S 44» 04 These are exhibited graphically in Fig. 12, and the method is as follows. Draw a straight line OX, and at suitable equal intervals mark points corresponding to the twelve months. Draw OY at right angles to OX, and with a suitable scale mark points corresponding to temperatures from 40° to 65°. For economy of space, temperatures below 40° are not marked. To save time and secure accuracy, specially ruled paper, called " squared paper ", is employed. At each of the 12 points on OX draw lines perpendicular to OY to represent the corresponding temperature as indicated on the scale on OY. These straight lines take the place of the columns in Fig. 11. This diagram shows at a glance not only the mean temperature for each month, but gives a clear picture of the rise and fall of average temperature tliroughout the year, the highest and the lowest. 63. A graph. It will be seen that when squared paper is used it is not necessary to draw the lines perpendicular to OX (as was done in Fig. 12). A point will mark the top of such a line, if it were drawn. We now join these points by straight lines as shown in Fig. 13. The succession of lines thus formed is called a graph. This is much more useful than the series of perpendicular lines in Fig. 12, and lends itself to important developments, as will be seen in later examples. Not only does it show more vividly the rise or fall each month, the highest and lowest temperatures, etc., but other features are apparent. For example, we get an idea of the average rate at which temperature is rising or falling during each month. This is indicated by the slope of the line corresponding to the month. Between January and February the slope of the GRAPHICAL REPRESENTATION OF QUANTITIES 91 line is very slight, the rise in temperature being very small. Between March and April the steep slope shows a rapid increase in temperature. The month of the sharpest rise IB 65 IB 60 55 50 45 40 rff 1 1 1 1 i 1 i 1 H 1 1 1 1 1 1 1 1 1 1 1 1 1 1 II 1 1 n 1 1 1 1 1 1 1 ft 1 1 1 1 1 1 II 1 1 B Jan Feb Mar Apl May Jne Jly Aug Sep Oct Nov Dec Fig. 13. is April, when the slope is greatest. The slope is reversed when the temperature falls, and the month when the fall in temperature is greatest is evidently November. 64. Examples of graphs and their uses. Graphs are employed in almost every branch of know- ledge, and a full treatment of them is impossible in a volume of this size. The following typical elementary graphs may serve to illustrate their nature and their uses. The first example is from Insurance. Example. The annual premiums charged by an Insurance Company for a policy of £100, at various ages are as follows: 9 2 TEACH YOURSELF ALGEBRA Age: ' 25. 30. 35. 40. 45. 60. Premium in £s. 2-33 2-50 2-91 3-31 3-81 4-63 The amounts of the premiums have been taken to the nearest new penny. The method of " plotting " these values, as it is called, is as follows. Y 4-5 4 M «*« .£3-5 E .2 3 E 2! o- 25 2> ii:! 1:138: +3 ^ :-s ftEft 7t ym i 5 r % Fig. 14. Two straight lines, OX, OY, are drawn at right angles (Fig. 14). Ages are marked off on OX, one small division representing a year. These are named as shown. Premiums arc marked on OY, one small division representing lOp. The lines OX and OY are called axes, OX being called the axis of x, and OY the axis of y, for reasons which will be apparent at later stages in the graphical work. GRAPHICAL REPRESENTATION OF QUANTITIES 93 As there is no premium less than £2, we begin marking the premiums by placing the mark for £2 at ; similarly on the x axis we begin the ages with twenty-five years at 0. By this means space on the paper is used to best advantage. In succession we find the points which show the premium corresponding to the ages given. These are marked with dots. The student should take squared paper and get these points for himself. The process of obtaining the points is called " plotting the points ". When this is done it will be seen that the points appear to lie on a regular, smooth curve, and this has been drawn on the figure. An important point now arises. If we are correct in assuming that points corresponding to the premiums for every fifth year lie on a regular smooth curve, then the values of these would seem to have been calculated according to a definite law or formula. If that is the case, then the premiums for the intervening years, which are not given in the table, would be calculated according to the same formula. It is therefore a reasonable deduction that the points corresponding to these. If plotted, would lie on the curve already obtained. Accordingly, if we note the intersection of the curve with the perpendicular line corresponding to any intervening year, the position of this point with reference to the scale on the y axis indicates the amount of the premium for the year selected. Thus, considering the point P, corresponding to the forty- tliird year, the premium as shown on the y axis is £3-60. Conversely, when the premium is £3 at Q, the correspond- ing age is thirty-six. This process of obtaining values from the graph which lie between those plotted is called Interpolation. 65. An example from Electricity. The table below shows the resistances in ohms for given lengths of wire of llie same material and cross-section. Length in yds. 100 120 170 220 Resistance in ohms. 2-5 3 4-25 6-5 94 TEACH YOURSELF ALGEBRA Draw the graph which shows the relation between resistances and length and find the resistance for a length of 200 yds. Two straight lines OX and OY are drawn as before (Fig. 15). 60 «50 £ 40 •ji 8 M ;i ;: i 100 150 200 250 Length in yards Fio. 15. Lengths are marked along OX, beginning with 100 yds. at 0, and each small division representing 5 yds. Resistances are marked along OY. five divisions repre- senting 1 ohm and .*. each division representing 0-2 ohm. Using corresponding values from the table, points are p.'otted as before. It will be seen that these points lie on a straight line. As in the preceding example, we deduce that a definite law connects resistance and length. Using the method of Interpolation we see that when the length is 200 yds. the resistance is 5 ohms. The graph being a straight line, it is easily produced as GRAPHICAL REPRESENTATION OF QUANTITIES 95 in the figure. This enables us to find the resistances corre- sponding to lengths beyond those given. For example, when the length is 250 yds. the resistance is a little greater than 6-2 ohms. This is called extrapolation. 66. An example from Mechanics. The following table shows tlie distances passed over in the times indicated by a body starting from rest. Draw a graph to show the relation between time and distance. Time in sees. . 1 2 3 4 5 Distance in ft. 2 8 18 32 60 60 tj so •8 £ 40 en 30 20 10 A 7 i! 1 2 3 4 5 6 A Time t in sees Fig. 16. Drawing the axes OX and OY, units are selected in accord- ance with the range of numbers to be represented (see Fig. 16). The points plotted are indicated by small dots. They g6 TEACH YOURSELF ALGEBRA appear to lie on a regular curve, and this is drawn to include all the points given in the table. As in previous examples, the regularity of the curve suggests that time and distance are connected by a definite formula. This will be familiar to those students who have studied Mechanics. The curve is known as a parabola. Reasoning as in the previous example we may use the curve for interpolation. The following examples illustrate this. Example I. Find the distance passed over in 3-6 sees. From the curve this appears to be 26 ft. The actual distance by formula is 25-92 sees. Exam pie 2. How long will the body take to travel 42 ft. ? Finding the point which marks 42 ft. on the y axis, and then looking for the corresponding point on the curve, this is seen, from the scale on the * axis, to represent 4-6 sees. The Mechanics formula gives 4-58 sees. Exercise 17. 1. The following table gives the values of exported manu- factured goods of a certain type in years as specificed. Year: 1930. 1931. 1932. 1933. 1934. 1935. Value in million £s. 60 6-2 4-9 5-1 4-8 6-6 Show the variations by means of a graph. 2. The temperatures between 10 a.m. and 8 p.m. on a certain day, taken every two hours, were as follows: Time: 10 a.m. 12 noon. 2 p.m 4 p.m. 6 p.m. 8 p.m. Temp. 47° 60° 64° 66° 49-5° 48" Exhibit these in the form of a graph and find what was the probable temperature at 5 p.m. 3. The expectation of life in years for males and females GRAPHICAL REPRESENTATION OF QUANTITIES 07 In this country at various ages is shown in the following table: Age: 20. 30. 40. 50. 60. 70. Expectation (Males) . 47-2 38-6 29-8 21-8 14-7 8-8 Expectation (Females) 60-3 41-6 32-8 24-4 16-7 10-2 From your graph estimate the expectation of life for males and females at (1) 35 years, (2) 55 years. 4. The average weight of boys of different ages is given In the following tables. Draw a graph to illustrate them. Age (years) : 11. 12. 13. 14. 15. Weight (lb.) . 80 85 92 101 114 From the graph find: (1) The probable average weight of a boy of 12$ years. (2) The probable average weight at 16 years. (3) If a boy of 13 J years weighs 93 lb., about how much is he below average ? 5. The rates of births and deaths per 1000 of the popula- tion of England and Wales in the years of census were as follows: Year: 1871. 1881. 1891. 1001. 1911. 1921. Births . 38-5 34-1 30-8 28-7 24-5 22-8 Deaths 22-3 19-7 19-7 17-2 13-8 12-4 Draw with the same axes the graphs of (1) the birth rate, (2) the death rate, (3) the excess of the birth rate over the death rate for the period. Can you from your graphs forecast the birth and death rates for 1931 ? 98 TEACH YOURSELF ALGEBRA 6. A train starts from rest, and its speed at intervals during the first minute is given in the following table: Time (sees.) 6 10 20 30 40 60 60 Speed (ft. per sec.) 8-5 14-6 23 29-2 33-6 37 39 Draw a graph showing the relation between time and speed. Does it appear that a definite law connects these ? What is your estimate of the speed after 23 sees. ? 7. In a certain machine the effort required to raise dif- ferent loads was found to be as follows. Draw a graph to show these. Load (in lb ) 8 12 16 20 30 40 Effort (in lb wt.) . 3 4-2 6-6 6-5 9-7 121 (1) Does the relation between load and effort appear to follow a definite law ? (2) What effort would be needed to lift 50 lb. ? 8. At the following draughts in sea-water a ship has the indicated displacements. Draught (in ft.). . 6-3 9 12 16 Displacement (in tons) S86 1018 1512 2098 What are the probable displacements when the draughts are 11 ft. and 13 ft., respectively? 9. The following prices of water turbines are taken from a maker's catalogue. Size of wheel (in ins.) IS 18 21 27 30 Price in la. . 68 80 94 130 153 Plot these, and from the graph estimate the price of a turbine with a 24-in. wheel. CHAPTER IX THE LAW OF A STRAIGHT LINE; CO-ORDINATES 67. The straight-line graph. Among the graphs considered in the previous chapter were some which were straight lines. This regular arrange- ment of the plotted points suggested that a definite law governed the relations between the quantities, corresponding values of which were the basis of the plotting. This is an important principle which calls for further investigation. The following example will serve as a starting-point. The proprietor of a restaurant calculated Uie following figures showing the connection between his net profits and the number of customers. Exhibit the connection by a graph. Number of Customers . 240 270 300 350 380 Net profit in £s. 0-50 2-00 3-50 600 7-50 Note. — By " net profit " is understood total receipts from customers, less expenses. Let * = the number of customers. ,, y = the net profit. Choosing two axes as usual, values of * will be plotted on the * axis, values of y on the y axis. When plotted as in Fig. 17, the points are found to lie on a straight line. Interpolation and extrapolation can be used as pre- viously, but two important questions arise: (1) What is the number of customers when there is no Profit? Zero on the profit (y) scale is shown where the graph cuts the * axis. Producing the graph it cuts at the point (A), where x = 230, i.e., when x = 230, y = 0. 99 ioo TEACH YOURSELF ALGEBRA (2) What happens when the number of customers is less than 230 ? If there is no profit when the number of customers is 230, there will be a loss when the number Is below 230. Let the straight line be produced below the x axis, as in Fig. 17. The amount of the loss, as with the profit, will be shown on the y axis, which must be produced. Fig. 17. The question of indicating loss as contrasted with profit now arises as it did in Chapter IV. We must proceed in the same way. The loss will be marked with negative numbers on the scale, as shown. It will then be seen that on the y axis there is thus constructed part of a complete number scale (see § 36). THE LAW OF A STRAIGHT LINE ioi From this scale, it appears that when x (the number of customers) is 180, y (the loss) is £"2-50. Problems will also arise when it will be necessary to use negative values of x, on the x axis produced. Consequently for a complete graph we shall require complete number scales on both axes, with zero common to both. In the particular example above, a negative number of customers would have no intelligible meaning. 68. The law represented by a straight-line graph. Our next step is to discover the nature of the law which a straight-line graph represents and how it can be formu- lated. We will use the above problem as our example. In that problem it was stated that the net profit was equal to the total receipts less the expenses. But the total receipts equal (the number of customers) x (the average amount paid by each). Let £a = the average amount thus paid. Then since * = the number of customers, ax = the total amount paid in pounds. Let £b = the expenses. Then net profits are (ax — b) pounds, i.e., y = ax — b. This gives the value of y in terms of a, x, and b, and is the form of the law connecting them. Of the four letters, a and 6 remain unchanged, while the number of customers (x) and consequently the profit (y) vary. .'. the law is not completely stated until we know the values of a and b. The method of doing this is suggested by example 2, §59. Two pairs of corresponding values of x and y can be obtained from the table of values or from the graph. For example, when when d(alg.) x = 320, y = 4-50 x = 250, y = 100. .2 TEACH YOURSELF ALGEBRA Substituting these in y = ax — b 4-5 = 320a — b . . 1 = 250 — b . . . g Subtracting (2) from (1) 3-5 = 70a and a = 005. Substituting in (1) we get b= 11-5. ,", the equation is y = 005*— 11-5. This Is the law of the straight line. If it is correct it must be satisfied by any corresponding pair of values of * and y, and this should be tested by the student. In particular if y = 0, i.e. there is no profit 005* - 11-5 = whence 005* =11-5 and * = 230. This agrees with the result found above. Thus it can be demonstrated that the equation y = 0-05* — 11-5 is satisfied by the co-ordinates of any point on the line and so is called the equation to the straight line. 69. Graph of an equation of the first degree. The equation which is represented by the straight line in the above problem (viz., y = 0-05* — 11-5) is of the first degree, that is, it contains no higher powers than the first of * and y. Two questions now suggest themselves: (1) Can every equation of the first degree in * and y be represented graphically by a straight line? (2) Conversely, can every straight line be represented algebraically by an equation of the first degree ? THE LAW OF A STRAIGHT LINE '°3 The answers to these questions will be apparent later, but for the present we will confine ourselves to drawing the graphs of some typical equations of the first degree. From these the answer to the first question may be deduced. Examples are given to illustrate the methods to be em- ployed. In all of these, since both positive and negative values of * and y will be involved, complete number scales will be used on both axes. Before calculating corresponding values of x and y, the student who is not familiar with the work should revise §50. 70. Worked examples. Example I. Draw (he graph of the equation: 2y — 4x = 3. This is not in the form used above, but it can readily be transformed thus: 2y - 4* = 3 1y = 4* + 3. y = 2x + 1-5. Giving suitable values to x, the corresponding values of y can be calculated, and so we get the following table: * -2 -1 1 1-8 2 y -2-5 -0-5 1 1-fi 3-5 6-1 5-6 A straight line is fixed by two points, but in drawing it from its equation three points should always be taken to check accuracy. In this example, as we are verifying the truth of a proposition, a number of points are taken, so as to make it clear that all such points lie on the straight line. The g^raph appears as shown in Fig. 18. This could be checked by finding for any point on the line the corresponding values of * anay. These values should satisfy the equation. Intercepts. It should he noted that when when 1) x = Q,y= 1-5, 2) y = 0, * = - 0-75. 104 TEACH YOURSELF ALGEBRA (1) 1-5 Is called the Intercept on the y axis (x = 0). (2) -0-75 x .. (y - 0). • It.- ... ■ ■••!,*•■ ■■■*•*•■ ^:::::::; :::::: IHIII jg iliiiiiiiHiinii (•■■■■■■■■•■■■■■■■■ii ■■•■■■■1 ■•■■ ■••■>• ■ ■ ■■■■■■■<■>•■ ■ inaafiiiia {!:{;: Hi ■■■ jijlljlijjiljllji ■■■ •■•■••■••■•■•••■§ :::::: ... ::::::::::::::::: :::::::::::::::::::: ■••*■■■■ ..:::. ■:: :::: ...Si: '<"■ ■■■■■■■■■■■■■■•■a .•n..::::. :::::.: :::::::::::!:::::::: Fir. 18. Example 2. Draw /Ae gra/>/i of the equation : 2x 4- y = 1. Transforming the equation we get : y = -2x + \ . . . (A) A table of corresponding values can be constructed as follows: THE LAW OF A STRAIGHT LINE '°5 1, . -2 -1 1 2 3 \f . 1 6 3 1 -1 -3 1 1 -6 With these values the straight line shown in Fig. 19 is drawn. For Intercepts on the axes when * = 0, y = 1 (inter- cept on y axis) when y = 0, x = 0-5 (inter- cept on x axis). It should be noted from this example that when the coefficient of * in the equa- tion arranged as in (A) is negative, the angle which the straight line makes with the x axis in an anti-clock- wise direction is greater than a rigjht angle. When the coefficient is positive, as in Fig. 18, this angle is less than a right angle. Fig. 19. Exercise 18. Draw the graphs which are represented by the following equations. In each case find the intercepts on (1) the y axis, (2) the x axis. 1. y = * + 2. 3. 4y = 6* — 5. 2. y = 1-5* - 1. 4. 3* + 2y = 6. 6. 2(* - 3) = My - 1). 7. The equation y = ax 4- b is satisfied by the following pairs of values of * and y: (1) *=l,y = 5 . (2) * = 2, y = 7. ,o6 TEACH YOURSELF ALGEBRA Find the values of a and b, and substitute in the equation. Draw the graph of this equation and find its intercept on the y axis. 8. A straight line makes an intercept of 2 on the y axis and of 4 on the x axis. Find its equation. 71. Position in a plane; co-ordinates. When a point is " plotted ", as in graphs previously con- sidered, its position on the graph was fixed by the corre- sponding values of x and y which were given in the table. For example, let P be a point such that in the table x = 2, y = 3. When plotted, the point appears as in Fig. 20, where : PQ is 2 units in length and parallel to OX, and PR is 3 units in length and parallel to OY. The intersection of these two straight lines fixes the position not only In the graph, but also relative to the axes OX and OY. The position of any other point can be similarly determined when its distances from the two axes are known. By such means might a man discover where, in a field, he had hidden a treasure — an act all too necessary in parts of Europe in these times. By remembering the distances OR and 00 along two boundaries of the field, he can " plot the point P " in the field. A little reflection will convince the student that the boundaries need not be at right angles to one another. The distances PQ and PR (Fig. 20) which thus fix the position of a point are called the co-ordinates of P with respect to the two axes. PQ, parallel to the x axis, Is called the x co-ordinate (or the abscissa). PR, parallel to the y axis, is called the y co-ordinate (or the ordinate). i;j;;liiE;ii;i;j;iili i iii: Ijj jjjjjii : : $:: I : j : r::::xj- i^jliflMilll;;;; I-ig. 20. THE LAW OF A STRAIGHT LINE 107 The notation employed to denote co-ordinates is (2, 3) or in general (x, y). The x co-ordinate is always placed first inside the bracket. Thus (5, 2) would represent the point in which x = 5, y = 2. In this way the positions of points relative to two axes can be described concisely. If, however, all points are to be included, complete number scales must be used on both axes, as shown in Fig. 21. In this way four divi- II ••; ym -- w" i iisy :;: -'II ffi i 1 H ] B -IX — 1 r:rri" m i CIS*. TT[T ' ' j ;.| :: ■■■ t •.'.—J : : !e- ; ::x- if §S|Mi 10 fffl :; 2:;:3::;4 :: '5" ;: : '■'- 1 a he •::::::::: :.. 1 ;;;: J'-: *? l| :' ;; ; E SY iMIllllll-^rLr-n-i-l^— iM Fig 21. sions, called quadrants, are found in the plane ; numbered I, II, III, IV as shown. The signs of the co-ordinates are regulated b^ the positive and negative parts of the number scales. The rule of signs is as follows : If (x, y) be the co-ordinates of a point : x Is +ve when measured to the right of 0. x is — ve „ „ .. left of 0. y is -f ve „ „ up from 0. y is — ve „ „ down from 0. ioS TEACH YOURSELF ALGEBRA As examples the co-ordinates of A, B, C, D in the figure are: A is (5, 4), B is (- 2, 4), C is (- 4, - 1), D is (3, - 2). Points on the axes. If a point is on the x axis the y co-ordinate Is zero. Thus E is (4, 0). If a point is on the y axis the x co-ordinate Is zero. Thus F is (0,-3). The intersection of the axes, 0, is called the origin. Since it lies on both axes, its co-ordinates are (0, 0). Thus the position of every point on a plane relative to two axes can be determined by co-ordinates. Latitude and longitude are a practical example of use of co-ordinates. They describe the position of a place with reference to the equator and the meridian through Greenwich as axes. The introduction of co-ordinates was due to Descartes, who published his book on Analytical Geometry in 1638. Exercise 19. 1. Write down the co-ordinates of the points in Fig. 22 marked A, B, C . . . G Fig. 2? THE L/>W OF A STRAIGHT LINE 109 2. Join OA and FC in Fig 22, and find the co-ordinates ,)f their point of intersection. 3. Plot the points (3, 1), (1. 3), (0. 3), (0. - 3), (4, 2-6), (-2.1), (-4. -2), (3,0). (-2,0). 4. Draw the straight lines joining the points (3, 1) and (1, 3) and also (— 2, 1) and (4, 2-G) as plotted in the previous question. What are the co-ordinates of the point of intersection of the two lines? 5. Plot the points (- 3, 2), (0, 2), (2, 2), (4. 2). What do you notice about these points? 6. Plot the points (3, 3), (1. 1), (- 1, - 1), (-2. —2). What do you notice about these points? 7. Draw a straight line through (3, 0) parallel to YOY'. What do you notice about the co-ordinates ol points on this line ? 72. A straight line as a locus. Let A (Fig. 23) be a point such that its co-ordinates are equal. Let (x,.y,) be its co-ordinates. Then y, = x,. Join OA and draw AP perpendicular to OX. Then AP = OP :. by Geometry /.OAP = LAOP since i-APO is a right angle. .-. each of the angles OAP, AOP is 45°. .*. A lies on a straight line passing through the origin and making 45° with the axis OX. Let B be any other point with equal co-ordinates (x t , yj so that BQ = OQ, i.e., y t = x v Then, for the same reasons as above, B also lies on a straight line passing through the origin and making 45° with the x axis. This must be the straight line OA , since only one straight line can pass through the origin and make an angle of 45° with the x axis. Similarly all other points with equal co-ordinates lie on the same straight line — i.e.. the straight line OA. produced, Is the locus of all points with equal co-ordinates. ho TEACH YOURSELF ALGEBRA These co-ordinates all satisfy the equation y = x, which Is therefore the equation of the line. This equation can be written in the form 2-1. X Fig. 23. .*, for any point sucli as .4, OP This ratio is constant for all points on the straight line, and is called the gradient of the line. A similar result holds for every straight line, consequently a straight line Is a graph which has a constant gradient. To include all points with equal co-ordinates the straight line must be produced into the opposite (3rd) quadrant. Then for any point on this line the co-ordinates are equal. THE LAW OF A STRAIGHT LINE m but they are both negative. Thus for the point C (Fig. 23) 2 these co-ordinates are (— 2, — 2) and the gradient is — ~, i.e., unity. The student who has learnt sufficient Geometry, or Trigonometry, will know, without the above demonstra- AP tion. that the ratio of ^ is constant lor any points. See also § 159. 73. Equation of any straight line passing through the origin. The conclusions reached above apply equally to all straight lines through the origin. The lines differ only in the gradient — i.e., in the value of ^. 4 1^4. rt i tut 2 3 ! ± + i u j tmIiIIiIttT it fr* 5 ■ .•::: :: : TititttP^ 1 -2 jfiSSfit : !lli I itttt t iiT' Hl^ J §5 E in i j£ tttt Cjj 1 1 1 1 1 1 h Fig. 2*. the ratio 112 TEACH YOURSELF ALGEBRA For example. i( the gradient is 2, then- = 2 and v = 2x. This is shown in Fig. 24. For any point P, on the line OP, is P0 - 4 -2 OQ ~ 2 - * Generally, if the gradient be denoted by m we have x y = mx. This is the general form of the equation of any straight line through the origin, wiiere m denotes the gradient. Negaclve gradient. If m is negative, the line will pass through the 2nd and 4th quadrants. Thus if »» = — 1 the equation is y= -x. The straight line is shown in Fig, 24. Considering any point A . the gradient is AB +2 . 0S--2- -1 - It should be noted that the angle made with the axis of x is 135°, angles being always measured in an anti-clockwise direction. 74. Graphs of straight lines not passing through the origin. In Fig. 25 the straight line AOB is the graph of y = x. If we now plot the graph of y = x -f- 2, it is evident that for any value of x the value of y in y = x + 2 is greater by 2 than the corresponding value of y in y — x. ;. the line for y = x + 2 must be parallel to y = x, but each point is two units higher in the^ scale. Thus in Fig. 25 the point A is raised to A', the origin to D. B to B' , etc. A'DB' therefore represents the graph of y = x + 2. The straight line y = x + 2 is the locus of all points whose co-ordinates are such that the y co-ordinate = the x co-ordinate 4- 2. It has the same gradient as y = x, but its intercept on the axis of y is -f 2. THE LAW OF A STRAIGHT LINE 113 Similarly the line y = x — ."5 is parallel to thi line y = x. with each point on it lowered by 3 units in the y scale. Evidently then we can generalise and state that y = x + b will always represent a straight line parallel to y — x and with an intercept of 6 units on the y axis for any value of b. HI! 1 TTT H ft £NA6t » 21; : 4 g mj jj Hiln |fe | : ""/: 1 ■ /^■V'-'* JintilM -■ 1; j ! B}. /n-i 1 -. 1 ■,•• / 1 iriiiiiiFi •/fill ffllj 111 — -—--}-— w p ^z~ /-r ' *' ::} •-'jltrii 7=t - • 9r* . i± 2; £ Tttr r nliSii -"5 *"*"H" 2 t£ 1 □ , Tttr- 1 ■ ? ITTT -^p:' Fig. 25. The same conclusions hold for lines with different gradients. For example, the equation y = 2x + b will always represent a straight line parallel to y = 2.t, •.«., having the same gradient, and with an intercept of b units on the y axis. Generalising, let m = the gradient of a straight line. Then y = mx -f- b always represents a straight line parallel to y = mx, I.e.. with the same gradient, and with an intercept of b units on the y axis. 114 TEACH YOURSELF ALGEBRA Examples. (1) y = 4# — 7 is a straight line whose gradient is 4, and whose intercept on the y axis is — 7. (2) y = 0-05* — 11-5 (see § 68) represents a straight line of gradient 005 and intercept on they axis equal to — 11-5. As shown in § 70, every equation of the first degree in two unknowns can be reduced to the form y = mx + b. We therefore conclude that the graph of every equation of the first degree in two unknowns Is a straight line. Further, it is evident that the equation Is satisfied by the co-ordinates of any point on the straight line. 75. Graphical solution of simultaneous equations. The conclusions reached in the previous paragraph can be used, as shown in the following example, to solve simul- taneous equations of the first degree. Example. Solve the equations: x + 2y = 5 3* - 2y = 7 (1) Draw tlie graph of x + 2y = 5. orainates is as follows : . . • • (1) • • . • (2) The table of co- * -1 2 6 y 2-5 3 1-5 Note.— When x = 0, y is the Intercept on the x axis; when y = 0, * is the Intercept on the x axis. It is useful to obtain these two points. The graph is the straight line marked A in Fig. 26. (2) Draw (he graph of Zx — 2y = 7. The table of values is : * 1 ! 4 y -3-6 -2 2-5 The graph is the straight line marked B. THE LAW OF A STRAIGHT LINE "5. Applying the conclusions, reached in § 74. Line A contains all those points whose co-ordinates satisfy the equation x + 2y = 5. Line B contains all those points whose co-ordinates satisfy the equation 3x — 2y = 7. • ••> ill ■■■■■■■■|H1 ■■■■■■■■■■.a :::::::; :r: ::::::::::: • ■■a .::::::::: ::::::::: >•■•■•••• •■■■>■•>•> v. ::::::: ■■«■«■■ ■••■■■ •■•«■• III- >■> • ■ ■ j a ■ ■■■■ ■ !■■■ ■ aaaa ■ill ill l;!i aaa! ■■■' ■■ *■>■•• ■ ■■■■ ■aaaa i:: ill! iij-' iff: ri: ::: :::::i!:ii:: i ■••■•••••< ■■■■•■■■■■a is::::i!!!i MM • >•• ■ •■■ iiii •■■•■■•■■ ::::::::: ?;{j ■••■•■ ■ mil ■ •Mil -'::ii: ;j;* • aaa ;:;: ■ -•a ■ aa* • ■* ■ aa 2 * I : ":: ,;;: Hljllilt j HWijil, ::• -:•:;:•: :ft tffft Fft fft tf+ m Q j: i ml | ■'/ W i .~ ■'rati :'.: *H U i4 ■ M« 1 :::: :: s .(■■•■■•■■I ■ ■■a :::: ::::! ••••••••• ••••aaaa* • •••■MM .... aaaa ■••■■ Fig. 26. There is one point, and one point only, whose co- ordinates satisfy both equations. That point is P, the intersection of the two graphs. The co-ordinates of P, by inspection, are (3, 1). .". the solution of the equation is x = 3.y= I. The student should compare the above conclusions with the algebraical treatment in § 55. Ii6 TEACH YOURSELF ALGEBRA Exercise 20. I. With the same axes draw the graphs of: (i)y = x. (2)y = 2x. (3)y=Jx. (4) y = - x. (5) y = - 2x. 2. With the same axes draw the graphs of: (1) y - *. (4) y = x - 1. (2) y = x + 1. (3) y = * + 3. 3. With the same axes draw the graphs of: U):V=i*- (2) y- 1*4-2. (3)y = i*-l. 4. With the same axes draw the graphs of : (2) y = 2x + 2. (4) y = ±* + 2. (2) v-2y = 4. (I) y = X + 2. (3) y = - * + 2. 6. Draw the graphs of: (11 2x+y = 3. (3) 5* 4- 2y = 10. (4) 4* - 5y = 10. In each case find the intercepts on the x axis and on the y axis. 6. Solve graphically the following pairs ot equations and check by algebraical solutions: (1) , + y = 7. K-J.-1. (3) 2x + 3y = 2. 4* + y = — 6. (2) 2* + y = 7. 3* — 5y = 4. 7. The straight line whose equation is y = ax — 1 passes through the point (2, 5). What is the value of a? What is the intercept on the y axis? 8. The straight line whose equation is y = 2x + b passes through the point (1, 3). What is the value of b? Draw the straight fine. What is its intercept on the y axis ? 9. The points (1, 1) and (2, 4) lie on the straight line whose equation is y = ax + b. Find a and b and write down the equation. What is the intercept of the straight line on the y axis ? (Hint.— See example 2 of § 59.) THE LAW OF A STRAIGHT LINE 117 10. Draw the straight line^ = 2x + 1. Draw the straight line parallel to it and passing through the point (0, 3). What is the equation of this line ? 11. Find the equation of the straight line passing through the points (1, 3) and ^2, yY What is the gradient of the line? 12. Find the equation of the straight line passing through the points (2, 4) and (3, 5). Find its gradient. CHAPTER X MULTIPLICATION OF ALGEBRAICAL EXPRESSIONS 76. (1) When one factor consists of one term. This has been considered in § 25, when it was shown that x(a -f- b) = xa + xb. 77. (2) Product terms. of binomial expressions — i.e., with two A y b D acq at] xb ab A typical example is {x + a) (y + 6). As in § 25, a geometrical illustration will help to make clear what is the product and how it is obtained. _ ,. In Fig. 27 ABCD is a rect- B angle with the sides (x -f a) and Si + b) units of length, and ivided to represent x, a, y, and b units of length. Lines are drawn parallel to the sides dividing the whole rectangle into - smaller ones whose areas repre- sent the products xy, xb, ay and Fig. 27. ab by means of their areas. The area of the whole rectangle = (x + a)(y + b) sq. units. Also the area of the whole rectangle = area of ADEF + area of EFBC = *(y + b)+a(y + b). .-. { x + a)(y + b)=x{y + b) + a(y + b) = xy + bx + ay + ab. This grouping suggests the method of multiplying (x + a)(y + o) algebraically. The second binomial (y + b) is multiplied in turn by each ix8 MULTIPLICATION OF ALGEBRAICAL EXPRESSIONS 119 term of the first factor. The sum of these is the final product. As the order in multiplication is immaterial so far as the final product is concerned, this could also have been obtained as follows. Writing the factors in the reverse order: (y + b)(x + a) =y(x + a) + b(x + a) = xy + ay + bx + ab. This is illustrated in Fig. 27. Worked examples. (a + b)(c + d) = a(c + d) + b(c + d) = ac -\- ad + be + bd. (x + S)(y + 3) = x(y + 3) + 5(y + 3) = xy + 3* + 6> + 15. (a + b)(c — d) = a(c — d) + b(c-d) = ac — ad + be — bd. {a — x)[b — y) = a(b — y) — x(b — y) — ab — ay — bx + xy. (x - 4)(y + 3) = %{y + 3) - 4(y + 3) = xy + 3* — 4y — 12. If the first terms in each factor are alike, the same method is followed, the product being simplified afterwards if necessary. Thus : (x + a){x + b) = x(x + b) + a(x + b) = x 9 + bx + ax + ab. This could be expressed as x* + {b + a)x + ab. This result suggests a quick way of obtaining the product mentally. The coefficient of x in the answer is the sum of a and b. The last term is their product. When a and b are numbers the sum and product will be evaluated, and the expression simplified. Examples. (x + 6)(* + 5) = *(* + 5) + 6(x + 5) - x* + 5x + 6x + 30 = *• + llx + 30. 120 TEACH YOURSELF ALGEBRA (a + 9)(a + 4) = a* + (9 + 4)a 4- 9 x 4 = a a 4 13a + 36. (x + 2)(x - 7) = *(* - 7) + 2(x - 7) = x* - 7* -(- 2* - 14 = x* + *(-7 + 2) - 14 = X s - 5x - 14. (a - 8)(a - 3) = a{a - 3) — 8(a - 3) = a 2 — 3a — 8a 4 24 = a 3 - 11a + 24. (x - 8)(x - 2) = x l + (- 8 - 2)x + (- 8) x (- 2) = x 2 - 10* + 16. (x - 8)(x + 2) = x* + (— 8 + 2)x + (- 8 x 2) = x* - 6* - 16. (x + 8)(x - 2) = x* + (+ 8 - 2)x + {+ 8 x (- 2)} = x» + 6x - 16. 78. When the coefficients of the first terms are not unity the rule still holds. Thus: (px + a)(qx + b) = px(qx + b) + a(qx + b) — pqx 2 + pbx + aqx + ab, which may be written as Pqx* -f (Pb + aq)x + ab. The form of the coefficients of x in the last line should be noted. It will be used later. Numerical examples of this form are common. The following illustrations show how to apply the rule quickly. Worked examples. Example I. (2x + 5)(3x + 4) = 2x(3x + 4) + 5(3* + 4) = 6x* + x(2 x 4 + 5 x 3) + 20 = 6x* + 23x + 20. The coefficient of the middle term can be obtained by multiplying as shown by the arrows below; then add the results: I (2x + 5)(3x + 4). MULTIPLICATION OF ALGEBRAICAL EXPRESSIONS 121 Example 2. (3* + 7)(2x + 1). -tut" Product = 6x* + *{(3 x 1) + (7 x 2)} + 7 = 6x* + 17x + 7. Example 3. &T -LJ- Product = 14x* + 4(7 x 3) - (5 x 2)} - 15 = 14x a + 11* — 15. Example 4. L-. (4x-7). A A Product = 12x* + *{(3 x - 7) 4- (- 2 X 4)} + 14 = 12x a - 29x + 14. If any difficulty is experienced the working should be set as shown in the second line of Example 1. 79. Multiplication of a trinomial. The method shown in § 77 may be usefully adapted to certain cases in which one of the factors is a trinomial. Example I. (x 4- 2)(x* — * + 1) = X ( X * _ x + i) + 2(x* - x 4 1) = X s - ** 4- x 4 2x a — 2x + 2 = X s 4 x* — x 4 2 (on collecting like terms). Example 2. (a 4 b)(a % - ab + b*) <= a(a* - ab + b*) + b{a* - ab 4 6*) = a 3 - a*b 4- ab* + a s b - ab* + b* = a 3 + b 3 (on collecting like terms). Exercise 21. Write down the following products: 1. a + x (b + y • 2. [c + d] («+/)■ 3. ax + b)(cy + d). 4. (a — x (b-y). b. x—y (a —b 6. (a - x (b 4- y). 7. a + x \b-v 8. (a 4- 2 (b 4 3 • 9. a-2 (6-3 • 10. (a-2 (b + 3). TEACH YOURSELF ALGEBRA 11. 13. 15. 17. 19. 21. 23. 25. 27. 29. (a + 2)(6-3). « + %. ab + 3). *-3). * + 3). x - 8y). x - 8y). (3* - 5y). 3x - 1). ..hi- 4y)- 3y (2x + 5y). 31. Simplify {(x + y)(a + b) - (ay + yb)) the result by x. 32. Simplify {(a + b)(a — c) + be} 4- a. 33. Find the value of (2x — y) (x + y) — (2x + y)(x — y) when x = 3, y = 2. 34. Simplify (1 - 2x)(l + 3.y) - (1 - 2y)(l + 3*) and find its value when * = 0-1, y = 0-2. 35. Find the following products: (ab + 6) (x - 10) (x - 10 \x-4y\ (x + 4y 3x-4y (2* — 3 1 + 3y (7* 12. 14. 16. 18. 20. 22. 24. 20. 28. 30. ;* + ?) x+ 10 x+ 10 6» + 8) fx-4y (a + 26 (4x+- (3x + (6x+l * + 5). (x + 3). (* - 3). d> ~ 12). (2a + 56). (7* 4 2). (3* - 4 . (3* - 5). (3a — 76) (6a -66). and divide (a) (x-y (6) (a + 2 M (1 + * (d) (x + a (x* + xy+y 2 ). (a* - 2a + 4). (1 _ x + X 2). (x* + lax + a 2 ). 80. Square of a binomial expression. In § 77 it was shown that (x + a)(x + b) = x* 4 x(a + b) + ab. Since this is true whatever the values of the letters, let b = a. Then (x + a)(x + a) = x* + x(a + a) + a x a = x l + 2xa + a*. (x + a) 2 = x 2 4- lax + a 2 . If a be replaced by — a throughout, then (x - a) 2 = x 2 - 2ox + a 2 . Geometrical illustration. Fig. 28 shows an illustration of the above result by means of a square whose side is x + a. MULTIPLICATION OF ALGEBRAICAL EXPRESSIONS 113 It is a modification of Fig. 27 and will be readily understood without further explanation. The student should draw a similar figure to illustrate (x - a) 2 = x 2 - 2a* + a 2 . Examples. x + 1 * = x* + 2x + 1. * — 1 * = x 1 — 2x + 1. a + 9 * = a* -f- 18a + 81. 1 - 5xy) 2 = 1 — 10*y 4 25* V- 2x + 7y) 2 = 4x 2 + 28xy + 49 v 2 . 3a — 106) 2 = 9a* - 60a6 + 1006'. ax a* X* ax x a Fig. 28. 81. Square of a trinomial. In the product (x + c) 2 = x* 4- 2cx + c 2 . Since * may have any value, replace it by a + 6. Then (a + 6 + c) 2 = (a 4 6) 2 + 2c(a + b) + c* = a 2 4- 2a6 + 6 2 + 2ac + 26c + c\ or, re-arranging, (a + 6 + c) 2 = a 2 4- 6 2 + c 1 + 2a6 4- 26c 4- 2ac. This may be stated in words as follows: The square of a trinomial is equal to (sum of squares of each term) 4 (twice the product of the three terms taken two at a time in every possible way). Examples. (a + b+ l)*=a* + b* + l+2ab + 2a + 26. (x-y + z)* - ** + (-?)■ + z> + 2 (* x -y) + 2 (**) + >(— 9 * ») = *' 4- y* + ** — 2xy 4- 2xx — 2yz. (a — b - c)* = a* + b* + c* — 2ab — 2ac + 26c. (x 4- y — 5) 1 = x* +y* + 25 - lOx - lOy 4- 2xy. Care should be taken in applying the laws of signs. I2 4 TEACH YOURSELF ALGEBRA 82. Cube of a binomial. (a + 6) 3 may be written as (a + 6)(a + b)(a + b) = (a + 6)(a 2 + 2«6 + 6 2 ). Multiplying these as shown in § 79 (a + 6) 8 = a(a % + 2ab + 6 2 ) + 6(a* + 2a6 + 6*) = a 8 + 2a 2 6 + ab* + a 2 6 + 2a6» + 6 s . ,\ (o + b) 3 = o 3 + 3a 2 b + 3ob 2 + b 8 (on collecting like terms). Similarly (o - b) 3 = o s - 3o 2 b f 3ob 2 - b\ Examples. (x + l) 3 = x> + 3** + 3* + 1. (* - l) 3 = x* - 3x» + 3x - 1. (1 + a 3 = 1 + 3a + 3a a + a 3 . (1 - a 3 = 1 - 3a + 3a 8 - a 3 . (2*+3y) 3 = (2x)» + {3(2x) 2 x 3y} + (3(2x) x (3y) 2 } + (3y)» = 8X 3 + 36**y + 54.*y 2 + 27y>. (x - 3y)» = r» - 9**y + 27xy* - 27y». Note.— In the cube of (x — y) the signs are alternately positive and negative. Exercise 22. Write down the following squares in full : 1. 3. 6. 7. 9. 11. 13. 15. (x + „.(. x + 2) 2 . a + 36) 2 . 2* + y) 2 - ab + 10) 3 . 4x + 5y) 2 . 5*y + 6)». 5*» + 3y 2 ) 2 . a + (* - 2) 2 . a - 36) 2 (x - 2y)«. (*y - 3) 2 . (Ax - by) 1 . (1 - 10x 2 )*. (3xy - 2y 2 ) 3 . 19. {(* + y) + 1}'. 21. (a + 6 - c) 2 . 23. (2* + 3y - 5z)». 2. 4. 6. 8. 10. 12. 14. - (B)' » s-«r- 20. {1 — (x - 2y)}». 22. (x - y + *) 2 . 24. (4a — 26 - 1)*. MULTIPLICATION OF ALGEBRAICAL EXPRESSIONS 125 Write down the following cubes in full: 25. (x+y)». 27. (a + 2) 3 . 29. (p + a) 3 . 31. (2*+y) 3 . 33. (3a - l) 8 . 35. Simplify the following: 36. Simplify: (1) (a + 6 j 26. (x-y)\ 28. (a-2) 3 . 30. (p - a) 3 . 32. (x — 2y) 3 . 34. (1 - 36)». ( X + y)* _ (** + _y*). - (a - 6|». - a + 6) 2 . 37. Simplify (x + 10)» - (s - 10) 2 . 38. If * = 3y + 1, express ** + ix + 4 in terms of y and find its value when y = 1 . 39. A square lawn the side of which is r ft. is surrounded by a path (Fig. 29) which is a ft. wide. Find an expression for the area of the path in terms of * and a. What is the area of the path when x = 30 ft. and a = 2 ft. ? 40. If the lawn of the previous ques- tion were a rectangle, x ft. by y ft. and the path were a ft. wide, find an ex- p 1G 29. pression for its area. " •-■at— \ •# 83. Product of sum and difference. If a and 6 be any two numbers, then the product of their sum and difference is expressed by (a + 6) (a 6). Using the method of § 77 to find the value of this (a + 6)(a - 6) = a(a - 6) + 6(a - 6) = a* — a6 + ab — 6*. .-. (0 +fa)(o - b) =0 2 - b 2 This important result can be expressed in words as follows : The product of the sum and difference of two numbers is equal to the difference of their squares. Examples. (x + 9)(* -9) =*»-81. (a6 + 10) (a6 - 10) -= a 2 6 2 - 100. 126 TEACH YOURSELF ALGEBRA (4* + 5) (4* - 5) = 16** - 25. (l+*)(l-*) = l-*«. (<*-*)(« + « = «•-*. (a b\(a b\ _ a* 6» \3 + 4/\3 4/ ~~ 9 16' i(a + b) + cM(a + b) - c\ = (a + b)* - e«. U + (b + c)H« - (6 + c)} = «» - (6 + «)*• 20. 21. a + x)(a - x). a + 26 (« -26). 2x + l)(2x — 1). l+« 2 ) (!-«*). Exercise 23. Write down the following products: 1. 3. 5. 7. 9. 11. 13. 14. 15. 16. 17. 18. 19. X t + yl )(x t_y2 ) _ 6. 8. 10. 12xy'+ \)(\2xy— 1). 12. (x + y) + z){{x + y) - z). a + x+y)(a + x — y). 2a + 36 + l)(2a + 36- 1). «-2> + 6)(x — 2y — 6). a + 2(6 + c)X« - 2(6 + c)}. 2* + 30- + z)K2* - 3(y + *)}. t* + «(* - 1). \2 + 3A2 _ 3/ - (* + a+|)(* + a-i). 2- (/• + ?)(/>-?)• 4. (4* + 3) ix — S (1 + 6%jil — tix). (2x* + 1 (2*> - 1). (3*y + 2 (3«y - 2). (ix - 7) Jx + 7). CHAPTER XI FACTORS 84. The work of this chapter will be the converse of thnt in the previous one. That was concerned with methods of obtaining the products of certain algebraical expressions. In this chapter we shall seek to find the factors of expres- sions of different types. A converse operation, in general, is more difficult than the direct one, and so it is in this case. By rules, on the whole simple and easily applied, the products of various kinds of factors are found. But in seeking to find the factors whose product produces a given expression, rules, even when they are formulated, are often long and tedious. In the main we rely on trial methods, which are, however, not haphazard but based on those rules by which the product was obtained. It is always possible to obtain a product when the factors are given, but we cannot always find factors for expressions. Most expressions have no factors, and so we can deal only with special types, such as were obtained as products in the work of the preceding chapter. 85. Monomial factors — i.e., factors consisting of one term only. This is the converse of the theorem stated in § 25. There we saw that x(o + b) = xa + xb. If we start with xa + xb and wish to factorise it, we see by inspection that x Is a factor of each term. It is there- fore a factor of the whole expression. To find the other factor we divide each term by x and add the quotients. As a result we get a + 6. xa + xb = x(o + 6). 127 128 TEACH YOURSELF ALGEBRA In finding the factors by inspection we are guided by a knowledge of the process by which when (a + b) is multi- plied by x the result is xa + xb. 86. Worked examples. Example I. Find the factors of 6a 2 + Zac. In this case we see that there is more than one factor common to each term. (1) 3 is the highest common factor of the numerical coefficients. (2) a is a factor of the other parts of each term. .". 3a is a factor of each term, and is therefore a factor of the whole expression. Dividing each term by it, 6a* -f- 3ac = 3a(2a -f c). Example 2. Factorise ox*y 2 — \Ox*y -f- 20y 2 . The highest numerical factor common to each term is 5, and the only other factor is y. :. 5*y — 10x*y + 20> 2 = 5y(**y - 2x* + iy). Exercise 24. Express the following as factors : 1. 6x + 12. 3. 4xy + 2y*. 5. 14*^ a — Ixy. 7. a* — ab + ac. 9. 15a 3 — 5a 2 b 4- 3a 2 i ! . 11. a*b + ab 1 — abc. 2. Zab + 2a. 4. 6a 2 - 4ab. 6. 16 — 32a 2 . 8. x 3 + 3% 2 — x. 10. 6a 2 c — loac 2 . . „ be 2 abc U - "6" - T' 13. Fill in the blank in the following: (7-4 x 13 1 ) + (7-4 x a 2 ) = 7-4( ). 14. Calculate the following as easily as you can : 18-6 2 + 18-6 x 1-4. FACTORS 129 87. Binomial factors. In § 77 we saw that (x + a)(y + b) = x(y + b) + a(y + b) = xy + bx + «y + ab. If we require the factors of xy + bx + ay + ab and similar expressions, we must work backwards through the steps shown above. The first step is to reach the stage x(y -\- b) + a(y + b). To obtain this the four terms of the expression must be suitably arranged in two pairs such that: (1) The terms in each pair have a common factor. (2) When this common factor is taken out, the same expression must be left In each pair. .". we group xy + bx + ay + ab in the form [xy -f- bx) + (ay + ab). Then taking out the common factor in each pair we get x(y + b) + a{y + b). Then y -f b is a factor of both parts of the expression and must therefore be a factor of the whole. Thus we get, on taking it out, ( y + b)(x + a). We will now apply this method to expressions of which the factors are not previously known, as they were in the example. 88. Worked examples. Example I. Find the factors of a 2 4- cd +• ad + ac. The first two terms have no common factor. Conse- quently the order of the terms could be changed in order to get two pairs, each with a common factor. .". we write a* + cd + ad + ac = (a 2 + ad) + (cd + ac). In this arrangement (1) a is common to the terms of the first group and c to those of the second. , 3 o TEACH YOURSELF ALGEBRA (2) the same expression a + d will be the other factor of each group. .-.we get a(a + d) + c(a + d). (a + d) being a factor of both parts is a factor of the whole. /. expression = (a -f d){a + c). Note. — Another possible arrangement was: (a* + ac) + (ad + cd) = a{a + c) + d{a + c) = ( + c)(a + d). There are always two possible ways of grouping. Example 2. Factorise, if possible, ab + ac + be + bd. This example is given to show the beginner where he might go astray. Hie expression can be grouped as a(b + c) + b(c -f- d). But the expressions in the brackets in the two parts are different. It is a bad mistake, which is sometimes made, to write down the factors as (b + c)(c + d)(a + b). Not one of these is a factor of the expression. Moreover their product would be an expression of the third degree. The given expression is of the second degree. On trying different groupings, it will be found that it is not possible to arrange in two groups, having the same factor in each group. There are no factors of this expression. Example 3. Find the factors of ab — 5a — 36 + 15. By arrangement into suitable pairs: ab — 5a — 36 4- 15 = (ab - 5a) - (36 - 15) = a(6-5 -3(6-5) = (6-5)(a-3). It should be carefully noted that when an expression is placed in brackets with minus sign in front, as was done with -36+15 above, the signs within the brackets must be changed. This is in agreement with the reverse rule given in § 26, cases (3) and (4). FACTORS Exercise 25. i3' Find the factors of: 1. ax + ay + bx + by. 3. ab — bd + ae — de. 5. x* + px + qx + pq. 7. ab + 5b + 6a + 30. 9. ab - 56 + 6a - 30. 11. ax* + a 3 x — ab — bx. 2. pc + qc + pd + qd. 4. ax — ex — ay J- cy. 6 x* — gx — hx -f gh. 8. a6 — 56 — 6a + 30. 10. 2a6 - 10a + 36 - 15. 12. x l + ax — bx — ab. 89. The form x* + ax + b. It was seen in § 77 that the product of two factors such as (x + 6)(x + 5) was x* + 11* + 30. Reversing the process, we must now consider how to find the factors of x % + 11* + 30. In the general case it was shown that (x + a)(x + 6) = x* + x{a + b) + ab. In this product (1) the coefficient of x is the sum of the numbers a and 6, (2) the term independent of x, i.e., ab, Ls the product of these numbers. Consequently in finding the factors of an expression such as x % + 11* + 30, we must find by trial two factors of 30 whose sum is + 11. They are + 5 and + 6. .'. the factors are (* + 5)(* + 6). 90. Worked examples. Example I. Factorise x* + 13* + 36. There are several pairs of factors of 36 — viz. (1 x 36), (2 x 18). (3 x 12), (4 x 91 and (6 x 6). We look for the pair whose sum is 4-13; this pair is 0. 4). V x«+ l3x + 36= (x + 4)(x + 9). Example 2. Factorise x* — 13* + 36. In this case the sign of the middle term is — and of the last term is +. i 3 i TEACH YOURSELF ALGEBRA .". we must look lor two negative factors ol 36 whose sum is - 13. These are — and — 4. x»- 13x4- 36= (x- 9)(x-4). Example 3. Facionse y l — 13y + 30. Proceeding as in the last example, we get : y* - IZy + 30 = (y - 10)(y - 3). Example 4. Factorise ** — 5* — 36. When the last term is negative the factors of 36 must have opposite signs. .'. the coefficient of x is the sum oi a positive and negative number, and the negative number must be the greater numerically, or we might say that 5 is the difference of the two factors numerically. The factors are clearly — 9 and + 4. x*-5x-36 = (x-9)(x • 4). Note. — The larger of the two numbers has the same sign as the middle term. Example 5. Facionse x % + 12* - 28. The two factors of 28 which differ by 12 are 14 and 2. They are of opposite signs, and their sum is -f 12. ,*. the factors required are 4- 14 and — 2. x* + 12* -28 = (x+ I4)(x- 2). Example 6. Factorise a 2 — 8ab — 486*. The introduction of the second letter makes no difference to the method followed, but 6 will appear in the second term of each factor. Thus a* - 8ab - 486* = (a - 126) (a + 46). Exercise 26. Find the factors oi: 1. ** 4- 3* + 2. 3. ** + 5* 4- 6. 5. x* + 7* 4- 6. 7. x* — I2xy + 20y*. 2. x* - 3x + 2. 4. x* - 5x 4- 6. 6. x* + 9* + 20. 8. a* — 15a6 + 366*. 9. x*y l + 15*y + 54. 11. y* - 2ly + 108. 13. x * _ x — 2. 15. x* + xy — 6y*. 17. 6* - 26 - 3. 19. x* + 13* - 48. 21. x*-xy- HOy*. 23. a* -a — 12. 25. p* - Mp - 72. 27. 1 — 8* — 20**. 29. />* + 4p — 45. FACTORS 10. 12. 14. 16. 18. 20. 22. 24. 26. 28. 30. '33 fl'6* - 19a6 + 48. x* — I2xy + 35y*. ** + * — 2. x t - X y- 6y*. b* + 2b- 3. x* — 13* - 48. a*- \U- 12. p* + p- 72. 1 - 9* + 20**. * a y* — 3xy — 88. p* + pq — 56o*. 91. The form ox* + bx + c. The factors of such an expression as this, where a, 6, c are numbers, will be of the form which occurred in the converse operation in § 78. These factors are best obtained by trial, as indicated in the following examples. Example I . Find the factors of 2** + 7* 4- 3. Write down possible pairs of factors systematically and find the middle term as shown in §78 until the correct coefficient of * is found. In this example the possibilities are : (1) (2* + 3)(* + l). (2) (2* 4- 1ft* + 3). In (1) the coefficient of * in the product is 5. „ (2) „ .. „ .. 7. The second pair is therefore the correct one -i.e., 2x* + 7x + 3 = (2x4- l)(x +-3). Example 2. Find the factors of 6** + 17* — 3. The minus sign of the last term indicates that the factors will have opposite signs in the second terms. This increases the number of possible pairs of factors. Among the possible pairs are : E (ALG.) 2* — 1)(3* + 3) or (2* + 1)(3* - 3). 6* —!)(* + 3) or (6* +!)(*- 3). 134 TEACH YOURSELF ALGEBRA Proceeding as shown in § 78, we find that the first pair of (2) is the one required, since the coefficient of * is (6 X 3) + (1 X - 1) = 18 - 1 = + 17. 6x* + I7x - 3 = (6x - l)(x + 3). Example 3. Find the factors of 4** — 17% — 15. The sign of the last term is again negative; therefore the factors will have opposite signs. Among the possibilities are: FACTORS '35 (2* + 5) 4* + 5) x — 3 x — 3 or (2* -5 or (4* — 5 (4* + 3) x — 5 or (4* - 3) x + 5). 2* + 3). x + 3). The first pair of (3) gives for the coefficient of x (4 x - 5) + (3 X 1) = - 20 + 3 = - 17. 4x*- I7x- 15= (4x + 3)(x-5). Exercise 27. Complete the following factors: 1. 3x*4- 10x + 8= (3* )(* ). 2. 12** — 17* + 6 = (4* ) 3x ). 3. 12** - 28* - 5 = (6* 2* ). 4. 9** + 43* - 10 = (9* ) * ). Find the factors of the following: 5. 7. 9. 11. 13. 15. 17. 19. 2** + 3* + 1. 2x* + 5* + 2. 4*» - 8* + 3. 6** — 11* + 3. 2a 8 + a — 1. 2a* — a — 6. 106* + 6-2. 12** + 5* — 2. 6. 3** — 4* 4- 1. 8. 6* 1 + 5* + 1. 10. 5** — 6* + 1. 12. 12** + 11* + 2. 14. 2a* — a— 1. 16. 106* — b — 2. 18. 8y* - Uy — 15. 20. 14c* - 17c - 6. 92. Squares of binomial expressions. A trinomial expression which is the square of a binomial can be recognised by applying the rule given in § 80. The standard forms are: and (a + 6)* = a* + 2a6 + ft* (a — b)* = a* - 2a6 + 6*. If the trinomial is a square, the following conditions must be satisfied: (1) The first and last terms, when the trinomial has been suitably arranged, are exact squares and positive. (2) The middle term must be ± (twice the product of the square roots of the first and third terms). Example I. Write ** + 6* + 9 as a square of a binomial expression. (1) The first and third terms are the squares of * and 3. (2) The middle term is + (2 X V** X V9) = 6*. x* + 6x 4- 9 = (x + 3)*. Similarly x» — 6x + 9 = (x — 3)». Example 2. Is 4** 4- 6* + 9 <» complete square? (1) The first and third terms are the squares of 2* and 3. (2) For a complete square the middle term should be 4- (2 X v^* 1 x V9) = 4- 12*. But the middle term is 4- 6*. .'. the expression Is not a complete square. Example 3. Is 4** — 20* 4- 25 a complete square? The middle term is — 2("V / 4**) X V25 = 2 x 2* X 5 = 20*. 4x* — 20x 4- 25 = (2x - 5)*. 93. Difference of two squares. In § 83 it was shown that (x + a)(x — a) = x*-«*. Conversely (x* — a*) = (x 4- o)(x — a). .'. an expression which is the difference of two squares has for its factors the sum and the difference of the numbers which are squared. Note.— No real factors can be found for the sum of two squares — i.e., ** + a*. i 3 6 TEACH YOURSELF ALGEBRA 94. Worked examples. Example I. Factorise 100x* — 1. The numbers squared are 10* and 1. ,\ the factors are the sum and difference of these — i.e., 100**- 1 = (I0x+ l)(IOx- I). Example 2. Factorise 36a*6* — 25. The numbers squared are 6ab and 5. 36a*6» - 25 = (6ob + 5) (6ab - 5). Example 3. Factorise (a + &)* — c*. Although one of the terms squared is replaced by a binomial expression, the rule above still applies. (« + &)» - c* = { (a + b)+ c\{(a + b) - c) = \a + b + c)(a + b-c). Example 4. Factorise {a + b)* — (c — a)*. (a + &)* - (c - a)* = {(a + b) + (c - a)}{(a + b) - (c - a)} = (a + 6 + c — a)(a + 6 — c + a) = (b + c)(2o + fc-c). 95. Evaluation of formulae. The above rule is frequently employed in transforming formulae and in arithmetical calculations. Example I . Find the value of 47-5* — 22-5*. 47-5* - 22-5* = (47-5 + 22-5) (47-5 - 22-5) = 70 X 25 = 1750. Example 2. Find Die area of a ring between two concen- tric circles of radii 9-7 in. and 8-3 in., respectively. The area of a circle is tv* where the radius is r. The area of the ring is the difference between the areas of the two circles. Difference in area = n x 9-7* — rt x 8-3* = ic(9-7 a — 8-3*) taking out common factor = *{(9-7 + 8-3) x (9-7 - 8-3)} = iiXl8xl-4 = 25-2-jt sq. Ins. FACTORS 137 Exercise 28. Write each of the following as the square of a binomial expression : 1. p* + 2pq + a*. 3. 9** -f 6* + 1. 5. x * + x + I 2. x* — 4xy + 4y*. 4. 16x* - 40*y + 25y*. „ a*,ab,b* 6-9 + 3 + 4. Express as complete squares: 7. (a + b)» + 4(a + b) + 4. 8. (x —y) 1 — \Q\x -y) + 25. Find the factors of the following: 9. x* - 100. 11. 4** - 9y*. 13. 121** - 36y». 15. 25 - \Gu". 17. 8a* - 506*. 19. 5** - 45y a . 21. (x + 2y) a - 16z a . 23. **—(>- + *)*. 25. {* - 8) a - 49. 10. a*6* - 25. 12. 25a' - 166». 14. 144/." - 169o*. 16. 1 - 225**. 18. 3*» - 75. 20. (a + 6)*-c*. HI -ft* 24. a a - (x - 2y) a . 26. (a + b)* — (a - b)*. Find the numerical values of the following : 27. 65 a - 35*. 28. 82 a - 68*. 29. 49* - 39*. 30. 24* - 18*. 31. 4-25* - 1-75*. 32. 17-5* - 12-5*. 22 33. Find the value of *(>-,* — r,*) when * = y, r, 12-5, r. = 8-5. v* — «* 34. If S = /=l-5. ~w find S when v = 14-5. « = 2-5 and 35. In the formula w = k(D* — d 1 ) find w. when k = 2-4, D = 8-5, d = 7-5. 36. If E = ml ~ v *- m2 ) find E when m = 24, v = 44, u = 16, g = 32. ■38 TEACH YOURSELF ALGEBRA 37. If s = $gt* find the difference between the values of s when t = 8 and / = 6. Take g = 32. 38. In the right-angled triangle (Fig. 30) a and b represent the lengths of the sides containing the right angle and c the length of the hypotenuse. We know from Geometry that c 3 = a* + b*. (1) Find a when c = 52, 6 = 48. (2) Find b when c = 65, a = 25. 96. Sum and differences of two cubes. In § 79, Example 2, it was shown that (a + b){a* - ab + b 3 ) = a 3 + b 3 . By using the same method we get: (a - b)(a* -{- ab + b*) = a* - b*. These results enable us to obtain factors of the expres- sions: a 3 + b 3 and a 3 — b 3 . Re-arranging them we have : a 3 + b 3 = (a + bUa* — ab + b*) . . (A) a 3 — b 3 = (a — b)(a* + ab + b 3 ) . . (B) It may aid the memory to note that (1) In (A), the sum of two cubes, the first factor is the sum of the numbers. (2) In (B), the difference of two cubes, the first factor is the difference of the numbers. The other factors differ only in the sign of ab. (1) When the first factor is the sum the sign is minus. (2) When the first factor is the difference the sign is plus. FACTORS 97. Worked examples. 139 X 9 + 1 = (X + 1 X 3 — 1 = (x - 1 1 - X 3 = (1 - X r* — X+l [X* + X+ 1) (1+X + X*). a* - 27 = (a - 3)(«» + 3« + 9). 8x* + 125 = (2x + 6)(4x» - 10* -f 25). Exercise 29. Find the factors ol: 1. 3. 5. 7 9. x 3 + c a . 1 + 8«». 8 + 27c». w> - 125»». 2. y* — a 3 . 4. x 3 — 64. 6. R 3 — 1. 8. #y> + I CHAPTER XII FRACTIONS 98. Algebraic fractions. In Algebra fractions have the same fundamental mean- ing as in Arithmetic, and arc subject to the same rules in operations with them. They differ only in the use of letters in the algebraic forms. An algebraic fraction is one in which the denominator is an algebraic expression. The numerator may or may not be algebraic in form. Thus - is an algebraic fraction, but ^ is not. The latter means £«, and thus J is merely a fractional coefficient. Similarly —j—l is an algebraic fraction but — g— is not. 99. Laws of fractions. It was stated above that algebraic fractions are subject to the same laws as arithmetical fractions. It is unneces- sary to state these laws here, but a few illustrations may serve to remind the student how they operate. It was pointed out in § 24 that a a x m I b x , a a — m and T = r— ; , b -7- m where m is any number. By means of this rule algebraic fractions can be simplified or reduced to lowest terms by dividing both numerator and denominator by the same number. This can be expressed by the phrase common factors are cancelled." This was assumed in §21. 100. Reduction of fractions. Examples were given in §24, but further instances are now given depending on the operations of the previous chapter. 140 FRACTIONS b '4' Example I. Simplify \_ bi r Using the rule of § 93, a + b a + b _ a*-b a ~ ( a + b)(a-b) = fTZTh ' on cancelling (a -f b)). Example 2. Simplify -* .+Jl=^?. Factorising numerator and denominator, *= - 4x- 12 _ (x— 2 ) {x ■ (!) x* + x-6 * (x-2)(x + 3) x + 3* The mistake is sometimes made of cancelling terms instead of factors. Thus, there can be no cancelling with the 3 and 6 in the above answer. Only factors of the whole expressions in numerator and denominator ran be cancelled. Factorising, the fraction becomes 3a(a — 2b)* 6a*(a + 5b)(a - 2bY The factors 3a (a — 2b) are common to numerator and denominator. Cancelling them, the fraction becomes, — 2b 2a (0 + 5b)* Exercise 30. Simplify the following iractions: 1. Zxyz G.v-z 2 ' -'. I5<?M» I 4 2 TEACH YOURSELF ALGEBRA 2x* + 2xy Axy — 4y r r fii'y + (i.vv 5 4*V + 4*^y»" 7 6. 3a 3 + 9a6 " 6« 3 6 — 6a 'S* a»_6* 9. ** — 6* + 6' a 3 + 6a«6 -f 9ab * a*6 + 5a6 1! + 66» - 8. a* — a6' x * — Q X + 20 x a + *- 20' 101. Multiplication and division. If necessary the student should revise § 24 and then pro- ceed to the harder examples now given. x x 2 Example I . Simplify ^j -4- ^TZTi' Factorising and inverting the second fraction : x (* + m» - 1) * + 1 • x« — 1 x + l Cancelling common factors we get: x- I Example 2. Simplify xt _ Q - — ^p-g — • Factorising the expression is equal to: x(x* - 27) . x 2 + 3x + 9 **-9 • * + 3 _ *(*-3)(*« + 8x + 9) _* + S_ (x-f 3)(*-3) ** + 3* + 9 = x (on cancelling factors). Exercise 31. Simplify the following fractions: L x* - 1 X y " 2. *y -r y -HI- FRACTIONS Q «' ~ 6 * . v ab + b * 4 3 - a* + 2ab + 6* x a* - a6" 2a» — 6a 8 6 a* + Sab + 2b* 5- 2a 8 - 2a6* X a - 36 ' 2x* + x — 1 . 2x* — x x — 2 • x % — hx + 6" a -f 2 8. 6 *'-*T 2 x 6. '43 a» — 49 . a + 7 a» — 9 - a + 3* 8* + 4 A 3% 2 - 5x + 2' 9. Simplify p-^Ti x 6 * ~ g ~ 4 and find its value when 6 = 5. 10. Simplify *-^^ X j4^t 8 and ** itS value when * = 1-5. 102. Addition and subtraction. The fundamental principles were examined in § 23. We now proceed to more difficult examples. Example I . Simplify ^-j - gt "_ p . Factorising, the expression becomes a a* a~=l (a + 6)(a-6) - The least common denominator is (a + 6) (a — 6). .". fraction a(a + 6) — a» ~(a + 6) (a -6) a* + ab — a * = (« + 6)(a-6) ab (a + b)(a-by Exam pie 2. Simplify — -j - j|_ft| . 144 TEACH YOURSELF ALGEBRA Factorising, the expression becomes 3 2a ; + b iT-b (a + 6)(a-6) _ 3(a + b) — (2a + b) _ 3a_+ 3ft - 2a — 6 = (a + b){a-b) (a + b)(a- b) a + 2b -(a + b)(o-bY This example is intended to remind the student of the note at the end of § 28 relative to a minus sign before a fraction. The numerator, as (2a -+- b) in the above example, should always be placed in a bracket on addition of the fraction, and the bracket removed afterwards. Example 3. Simplify x-l — x — 2 x* + 4* + 3' x + 2 Factorising, the fraction becomes x-l x+2 ~ (x- 2)[x + 1) (x + 3)(*+l) (x - l)(x + 3) - (x + 2)(x - 2) (*+l)(*-2)(* + 3) ( X * + 2x - 3) - (*» - 4) - (x+l)(x-2)(*-f 3) _ x t + 2x — 3 — x* + 4: = (*4-l)(x-2)(*-f3) 2x4- I -(x+l)(x-2)(x+"3j- Example 4. Simplify r. * x x x _ . x* — 1 — l = x*-r\- x - r —x— X X X ■X X x * _ i - j*ZT]- FRACTIONS '45 Example 5. Convert lite Jormula «a« — ^- into one in which the subject is R. 11 1 R R, R, i.e., 1 _ R t — R 1 R~ R t R t ' ••• «-A 0— *«!-» »*~4 Note. — The student should not make the mistake of inverting at the outset. The fractions on the right side must be added first. Exercise 32. Simplify the following: * x —y 3.„-^U+ 3 2. 3(a-b)^2(a + bY 6.3 2 ! x — 1 4.1-1 + 3 1 7. "^3 T (*-3)«- 3 4 1 _ a I" (l _ fl )»- 9 * * + 2>. h 11 -A- * 8. 8. * ' x + 2" 2* 3* 3(* + 1) 2(* - 2) - * + 3 x-3 x — 3 x + 3' 13. 14. 1 • + : ' 10 -srri-5+ 1 - a + ft * + ^" *+y * — y x y x* — y*- x*—y* (x — y)* x + y Transform the following formulae to those in which the subject is/?: IK 1 1 -L 1 1 9 3 1R 1 _ 3 2 i 4 6 l9I, * = r + s letters. » « P-ph TEACH YOURSELF ALGEBRA 1 1 r — s 3 find i? in terms of the other P+9 P» J. 0* 21. Simplify? + 0-y±-^- 22. Simplify y±i£ - find the value of 5P. ' -P. 23. If i = i- + i- + ^ findi? when /?, = 8-6, R t = 4-3. /< Kj K t K 3 and i?3 = 3. 24. Given «(P — \Q) = 6((? — |P) rearrange the terms so as to express P in terms of the other quantities. 25. Given J = -st—, find n in terms of the other R + nr quantities. Find n when / = 2, E = 1-8, R = 2-4, r = 0-5. 103. Simple equations Involving algebraical fractions. The following examples will illustrate the methods employed. Example I. Solve the equation : 3 5 x — 2 x— 1" The least common denominator is (x — 2) (x — 1). We therefore multiply both sides by this. Then 3(x-2)(x-l) _ 5(*-2)(x-l) inen x-2 - x-1 Cancelling 3(x - 1) = 5(x - 2) or 3x - 3 = 5x - 10. 2x = 7 and x = 3-5. Example 2. Solve for n the equation: »-2 T n-3 n Multiply throughout by the least common denominator «(» — 2)(« — 3). FRACTIONS 147 Then n(n - 3) + n(» - 2) = 2(n - 2) (« - 3). n » _ 3n + n* — 2n = 2(n* - 5» + 6) or 2n* — 5w = 2n* — 10n + 12. The terms involving 2»* disappear. 5n = 12 12 and n= T . Exercise 33. Solve the following equations: l.P-l=2. ** 2x 3.^ = 3. x — 2 6. 0-8 = 15n 4 + l-4n 3/> + 23 _4 • 3p + 12 _ 3' 9. 2. 2*- 1" = 5. 4. 1 —X x + l~ = 4. 5 4 b 2x + 5 1-8 X + 5" 8 X 1 '-2 4' x 5 (*-2)(*+l)- x— 2 x+1 10.?-- * =0. x * — 2 11. In the formula -> = find v when/= 8, « = 2. / v u J 12. For what value of n is . ~ „ ■ equal to i ? 4W + 13. If C = ^find R when C = 7-5, V = 60. 14. If r = R ( £ ~^ find K in terms of r, R, E. CHAPTER XIII GRAPHS OF QUADRATIC FUNCTIONS 104. Constants and variables. In the formulae which have been considered in earlier chapters it will be seen that the letters employed represent two different kinds of numbers. (1) Some represent constant numbers — i.e., numbers which remain unchanged in the varying cases to which the formula applies. (2) Others represent variable quantities. Example I. Consider the formula for the circumference of a circle, viz. C = 2w, where C = the length of the circumference, r = „ „ radius. Two of these four symbols represent constants. (1) The number 2 (which is fixed for any circle), (2) The number k (which has always the same value in any formula in which it is used). It represents the constant ratio of the circumference of any circle to its diameter, or the area of a circle to the square of its radius. Two of the letters In the formula are variables, they are different in value for different circles, the variation of C depending on changes in r. Example 2. In § 68 it was shown that the formula, or equation connecting profit and customers, was y = 0-05* - 11-5. It was pointed out that while the profit represented by y depended on the varying number of customers x, the numbers 0-05 and 11-5 remained constant, representing GRAPHS OF QUADRATIC FUNCTIONS 149 respectively the average amount paid by each customer, and the fixed charges. Example 3. This last example was a special case of the equation to a straight line, which in § 74 was shown to be represented in general by the equation: y = mx -f b. For a particular straight line, m is constant, representing the gradient of the line, while b is the fixed distance inter- cepted on the y axis. But x and y vary for different points on the line. They represent the co-ordinates of any point. 105. Dependent and independent variables. The variables are seen to be of two kinds. Considering the case of the circumference of the circle, the length of this depends on the length of the radius. In the second example the profit depends on the number of customers. A variable which thus depends on another variable for its value is called a dependent variable. The other variable upon which the first one depends is called the Independent variable. As another example, if a motor is moving with uniform — i.e., constant — speed, the distance travelled depends on the time. Thus the velocity is a constant, time is an independent variable, distance is a dependent variable. Again the cost of a quantity of tea depends on the weight bought, the price per lb. being constant. Thus weight is the independent variable, cost „ dependent variable. Graphs. In plotting graphs which show how one quantity varies as another varies, the independent variable Is always measured on the x axis and the dependent on the y axis. 106. Functions. When two quantities are connected as shown above, the dependent variable Is said to be a function of the independent variable. Thus: 15° TEACH YOURSELF ALGEBRA The circumference of a circle is a function of its radius. The area of a square is a function of its side. The distance travelled by a car moving uniformly is a function of the time. If a spring be stretched by a force, the extension of the spring is a function of the force. Generally if a quantity denoted by y depends for its value on another quantity x then y is a function of x. Thus in each of the following examples y = 2x y = x» y = 3(*-l)(*-3) the value of y depends on the value of x. If any particular value be given to x, a corresponding value of y can be calculated. In all such cases y is a function of x. The idea of a function is probably the most important in modern Mathematics. From the above examples the follow- ing definition can be deduced : Function. If a quantity y is related to a quantity x so that for every value which might be assigned to x there is a corresponding value of y, then y is a function of x. Thus for every length that may be chosen for the radius of a circle, there is always a corresponding length of the circumference. If a man works at a fixed rate per hour, then for any number of hours that he works, there is always a corre- sponding amount of pay. His pay is a function of the time he works; the rate per hour is a constant. Again if y = 2X 3 , then for every value which we may choose to give to *, there is a corresponding value of y. Consequently, y is a function of x. 107. Graph of a function. If y is a function of x, and since, by the above definition, for every value assigned to x there is always a corresponding value of y, then these pairs of values of x and y can be plotted, and the assemblage of points so plotted will be the graph of the function. GRAPHS OF QUADRATIC FUNCTIONS 151 Thus every function has a distinctive graph, which will be a regular curve or a straight line, and by which it can be identified. If the function is of the first degree, and does not involve any algebraical fraction, then, as we have already seen, the graph will be a straight line. For this reason a function of the first degree, of which the general form is y = mx -f- b, is called a linear function. If, however, the function involves a higher power of *, such as x % , x 3 , etc., or involves an algebraic fraction such as y = z< tne graph will be a regular curve, the shape of which will differ with the nature of the function. 108. Graph of a function of second degree. The simplest form of a function of the second degree is that which is expressed by y = x 2 . This is called a quadratic function, from the Latin quadratus (squared). The area of a circle, A = rcr 2 , is a special form of this. To plot the curve of y = x 2 . We first assign values to x, calculate the corresponding values of x 2 , or y, and tabulate them as follows: * -3 -2-5 -2 —1-8 — I 1 1-5 2 2-25 3 y 9 0-25 4 2-25 1 I 2-25 4 6-2.5 9 As wide a range of values is taken as the size of the paper will allow to be plotted. Since the values of y increase more rapidly than those of x, more room is needed on the y axis, but as the square of a number is always positive, no negative values of y are necessary. The * axis is therefore drawn near the bottom of the paper, as shown in Fig. 31. Selection of scales on axes. The scale chosen must depend on the number of values we wish to include within the limits of the paper. The same scales need not be chosen on the two axes, and in this particular curve it will be better to take a smaller scale for values of y, because their values increase more rapidly than those of x. It must be remembered, however, that the true shape i 5 2 TEACH YOURSELF ALGEBRA of the curve will be shown only when the same units are taken on both axes. The symmetry of the values of y, as shown in the table, suggests that they axis should be in the middle of the paper. The points when plotted appear to lie on a smooth curve as shown in Fig. 31. •■*■■••■• ......I. ..•«••■■ ::::: :::::•■ ;:||P| ■■■■■■■■■■■■■■■■■> ■ ■■•■•■■■■■■finii P 1 SB £ ;!=!£ ========== ■■■■■■■IIIMII"" ■■■■•••"•■■J ;■:: ■•••■■■■■>• ■■■■■■■■■■a ■ ■■»■>■■■■ ■ ■■IIIIUII •>■■■■-■*■ ■■■•■■•■■■■ ■■■■■■■■■■■■■■•■■■i ■«■■•■■■•■••■ 1 flilllllHHI •••■■•iiiiiinKiii ■ ••••IIIM» ■■■■■■ - — . TTT T fir* - ir-" iV ig$ ::::■ TTT T ~\ MTr ■ t ■■■ <*«■■■■■ ■■•■ •■■■••• :::::.-:::: ■•••••■.•■• miiiiiiniiiiHH ■■■■■■■•■ •■■■>••>•■ •■■■•-•-••■■•■•a** ..<■••>•■>•■■■•>••• ■■■■■■■■■••■■■■•■•i m •)■■■••*■■■ V ■■■■■■■■■■a :::::::i::::::lgji SHE •::: m ■:■:■:£:: :::& :^::::::::Ki:S: ■■■•■■•■*■■ ■a ■••■■■■■■•>■ a aaaa •■•■ >■ ■>■■■■•■ ■•■■■ Fig 31. The curve is called a parabola. When inverted it is the curve described by a projectile such as a shot from a gun, or a rocket when fired into the air. 109. Some properties of the curve of y = x 3 . The following points should be noted about this curve: (1) The symmetry of the curve. Positive and nega- tive values of x produce equal values for y. If there- fore the curve were folded about the y axis, the two parts of the curve, point by point, would coincide. The curve is therefore said to be symmetrical about OV, and OY is called the axis of symmetry. GRAPHS OF QUADRATIC FUNCTIONS 153 (2) The minimum value of the curve is at the origin. The curve is said to have a turning-point at the origin. (3) The slope of the curve is not constant, as with a straight line, but increases from point to point as x increases. The gradient is clearly a function of x, since its value depends on the value of *. (4) The curve may be used to read off the square of any number within the range of plotted values, and also, conversely, to determine square roots. Thus to find V3, take the point on the curve corresponding to 3 on the y axis. It is then seen that there are two points on the x axis which correspond to this, the values of % (i.e., V3) being +1-73 and —1-73 (see § 43). 1 10. The curve of y = — x*. All values of y for this curve are equal numerically to the corresponding values of y in y = x*. but are negative. The shape of the curve will be the same, but inverted, as in Fig. 32. I'ig. 32. 154 TEACH YOURSELF ALGEBRA All the values of y being negative, the x axis is drawn toward the top of the paper. The curve has a maximum point — viz. zero — at the origin. From this point the curve shows the path of a bomb dropped from an aeroplane at — ignoring air resistance. III. The curves of y = ox J . The curves represented by y = ax*, where a is any num- ber, are all parabolas differing from y = x* only in having different slopes. Fig. 33. Considering y = 2x* as an example the table of values for plotting the curve arc found in the same way as for y = x % This curve, as well as that of y = \x*. are shown in Fig. ^ Y* 33, and contrasted with y = x If a be negative, we get a corresponding set of curves similar to y = — x 1 , as in Fig. 32. GRAPHS OF QUADRATIC FUNCTIONS 153 1 12. The curves of y = x 3 ± o, where is any number. The curve of y = x % + 2 is evidently related to that of y = x* in the same way that y = x + 2 is connected with y = x (see § 74). Each ordinate of x l + 2 is greater by 2 than the corresponding ordinate of x 2 . Fig. 34. The curve of y = ** + 2 is therefore the same as y = x* raised two units on the y axis. It appears as shown in Fig. 34. Similarly, the curve y = x* — 3 is the curve of y = X s , but every point is three units lower than y = x* for corresponding values of x. Generally the curves represented by the equation y = ** ± a are a set of curves similar to y = x* and raised or lowered by an amount equal to ± a, according to the sign of a. ■56 TEACH YOURSELF ALGEBRA Similarly, the curve of y = — x* gives rise to a set of curves included in the equation y = — x* ± a. 1 13. Change of axis. On examination of Fig. 34. it will be seen that the curve of y = x l + 2 differs from the curve of y = x* only in the position of the x axis. For the curve of y = x % + 2 the Fig. 35. x axis is two units below the lowest point of the curve. Consequently the curve of y = x % can be changed into that of y = x* + 2 by drawing a new x axis two units below the original, as shown in Fig. 35, and re-numbering the y axis. Similarly, y = x* becomes y = x* — 3 by drawing a new x axis three units above the original, as shown, and again re-numbering the units on the y axis. GRAPHS OF QUADRATIC FUNCTIONS 157 114. Curve of y = (x - l) a . The table below shows the values of (1) y = (x — 1)* and (2) y = x 2 , for corresponding values of x. X -3 -2 -1 1 2 3 4 x* . 9 4 1 1 4 9 18 (* - !)• . | 16 4 1 1 4 9 :::::ir.::::i ::«::.•::::::: ::m ;::::::::: •■•a* •■•■■••■•■ ■■■■■■■••*■■«•- ■■■■■■■■■■•■•■a mi r fffnff' :■:■!!!:!•: BHaSSffESaBi • ■•■•••• .•• « ■•■■refill •■■•■•■•■■at SSrffi^ ■■•■■■■■■■■■■■I • • ■■•■* Biiiiiii! •■■■■a ■■••■■•■*i ■ >•-*, ■■■■■■•2, ■ -■>■•'■*■■■■»' • •••MlMIIMJl ■•■■•*■■■■■■■ ■-»•«.«-! .•■. a ■■■>■>*( 1 •■ ■ ■•■>»••> ■•■>■ ::;.:::;::;;::; L it ftt p f m? fftnm 3 ••■■•••••■a Fig. 36. A comparison of the sets of values of two functions shows that both have the same sequence of values, but those of (x — 1)* are those of x* moved one place to the right in the table for consecutive values of x. Consequently the curve of (x — l) 2 must be the same as that of x % , but moved one unit to the right. Therefore, as in the previous paragraph, but moving the y axis instead of the * axis, the curve of x 1 will be changed into the curve of (z — 1)* by moving the y axis ■58 TEACH YOURSELF ALGEBRA one unit to the left and re-numbering the * axis (see Fig. 36). Similarly if the y axis be moved two units to the left, we get the curve of (x — 2)*. Or if the y axis be moved one unit to the right we should have the curve of y = (x + 1)*, and so for similar functions. IIS. Graph of y= (x - I) 2 -4. By combining the operations illustrated in the preceding two paragraphs — i.e., by moving both axes — the curve of - : ' !| ! l l l li l l lll l l lllllll ll l 'TI!iii!5 i *-* tr r f J T TTT ' ] 'k", ■ _ J.Q ill t *X i ! ■h* *H : i : 1 fi Bin ±ii lit : i it i 111111 i±sl~ ill I liffllfiilil I ii Fig. 37. y = x 2 can be changed into the curve of such a function as y = (x — I) 2 — 4. This can be done in two steps: (1) By moving the y axis one unit to the left we get the curve of y = (x — 1)*, as in § 114. (2) By now moving the x axis 4 units up the y axis the curve of y = (x — l) 2 becomes y = (x — 1)* — 4. With these two new axes, both re-numbered, the curve is as shown in Fig. 37. 1 GRAPHS OF QUADRATIC FUNCTIONS '59 In this diagram the new axes have been drawn and the new numbers on the axes marked in larger figures. The expression may be simplified, since (x — l) 2 — 4 = x* — 2x + 1 — 4 = x* - 2x - 3. The curve in Fig. 37 Is therefore the graph of y = x* — Ix — 3. We shall see later (see § 1 21) that every quadratic function of x can be reduced to such a form as this, or the simpler forms of §§113 and 114. We may therefore draw the conclusion that: The graph of every quadratic function of x is a modified form of y = x 2 and is therefore a parabola. 1 16. The graph y = x 2 — 2x — 3. We have seen in the preceding paragraph that this graph can be obtained from the curve of y = x a by changing axes. In practice, however, the method given below will be found more convenient in most cases and usually more accurate in practical application. A table of values is constructed as follows: X -3 -2 -1 1 2 j 3 4 x' . 9 4 1 ' 1 4 9 16 -2*. + 6 + 4 +2 | -2 —4 -6 —8 -3 . -3 -3 | -3 -3 -3 -3 -3 -3 y ■ 12 6 -3 -4 -3 5 Note. — Be careful not to add the value of x in each column. It is well to draw a thick line under the values as shown, to remind one when adding. From the values in this table the student should plot the curve and compare the figure obtained with that in Fig. 37, arrived at by change of axes. It will be seen that the minimum value of x 1 — 2x — 3 is — 4, when x = 1, and the line x = 1, perpendicular to the x axis, is an axis of symmetry. i6o TEACH YOURSELF ALGEBRA 1 17. Solution of the equation x 3 — 2x — 3 = from the graph. At the points where the curve of x* — 2x — 3, in Fig. 37, cuts the axis of x, the value of y is zero — i.e. : x 2 — 2x — 3 = 0. The values of x at these two points arc: x = 3 and z = — 1. We have thus found two values of x which satisfy the equation x* — 2x — 3 = 0. The equation has therefore been solved by means of the graph. 118. Graph of y = 2x s - 3x - 5. This is a slightly more difficult graph to plot. The table of values is as follows : X -3 -2 -1 i i 2 3 4 S 2x' . 18 8 2 1 2 8 18 32 60 -3x. 6 3 -U -3 -6 -9 -12 -16 —5 . -5 -5 -5 .-1 -5 -5 —5 -5 -5 -6 y ■ 22 9 -6 -e -6 -3 4 16 30 In this example a wider range of values is shown. As the values of y increase rapidly smaller units are taken on the two axes. The graph is as shown in Fig. 38. The lowest point, N, giving the minimum value of the expression corresponds to x = J. From the table of values this point is seen to be half-way between x = J and x — 1, since these values of x give the same values of y — viz., — 6. The minimum value, represented by MN, is — 6J. The ordinate through M, of which MN is part, is the axis of symmetry of the curve. The solution of the equation : 2*» _ 3x - 5 = 0, GRAPHS OF QUADRATIC FUNCTIONS 161 will be given by the values of *, where y = — i.e., at A and B, where the curve cuts the axes. These points give: *— — I, or 2-5. Fig. 38. 1 19 Graph of y = 12 - x - x a . In this example the coefficient of x* Is negative. Conse- quently the curve will take the form ol y = v 2 , as shown in Fig. 32 — i.e., it will be an inverted parabola. The table of values is as follows : 162 TEACH YOURSELF ALGEBRA * —6 -4 -3 -2 -1 | 1 2 3 4 12 . 12 12 12 12 12 12 12 12 12 12 — x . 5 4 3 2 1 -1 -2 -3 -4 -X* -25 -18 -9 -4 -1 -1 —4 -9 -16 y ■ -8 01 6 10 12 12 10 6 -8 The values of y show the symmetry of the curve. The highest point, giving the maximum value, is seen to be JBr half-way between x = — 1, and * = — i.e., where x =» — $. The curve is as shown in Fig. 39. The maximum value is at M, where the value of y is 11 J. The ordinate from M — i.e., MN — is the axis of symmetry for the curve. We can obtain from the curve the solution of the equation : 12 — * — ** = 0—i.e. x* + x — 12 = 0. GRAPHS OF QUADRATIC FUNCTIONS '63 This will be given by the values of x at A and B, where the curve cuts the x axis. The solutions are: x = 3. x = — 4. Exercise 34. 1. Draw the curve of y = x % between the values x = + 3 and x = — 3, taking the units as large as possible. From the curve write down the values of: (1) 2-3». (2) Vf. (3) V3-5. 2. Draw the curve of y = £x a between x = + 4 and x = — 4. Find from the curve the values of x such that: (1) i*» = 0-8. (2) \x* = 2. (3) x* = 12. 3. Draw the curve of — J* a between x = -4- 4 and x — 4. From the curve find the values of : (1) - J*» = - 3. (2) - x* = - 5. (3) 2. 4. Draw the curve of y = x 1 and by change of axes obtain the curve of y = x a -\- 3. From the curve find the values of x such that : (1) x* + 3 = 5. (2) x* + 3 = 9. 5. Draw the curve of y = x 2 and by change of axes obtain the curves of: (a) y=(x- 2)*. (b) y=(x + 3)*. From the curves find the values of * such that: (1) (x - 2)» = 3. (2) (x + 3)» = 1. (3) (x - 2)» - 5 = 0. 6. Draw the curve of y = (x + 2)* — 2 by means of the curve of y = x* and changes of axes. Use the curve to find the values of x when: (1) (x + 2)* - 2 = 0. (2) (x -f 2)» - 1 = 0. (3) (* + 2) s = 8. 7. Draw the curve of y = x* — 6* -f- 6. Find the least value of this function and the corresponding value of *. Use the curve to find the values of x when: (1) x* — 6* + 5 = 0. (2) x* — 6x + 6 = 6. 164 TEACH YOURSELF ALGEBRA 8. Draw the curve of y = x % — 4x -f- 2. Find the minimum value of the function and the corresponding value of x. Use the curve to solve the following equations: (1) x* — 4* — 2 = 0. (2) x* — 4x — 2 = 3. 9. Draw the curve of y = 2x* — 5x + 2. Find the minimum value of the function and the corresponding value of x. From the curve solve the equations : (1) 2x* — 5x + 2 = 0. (2) 2x* — 5x — 1 = 0. 10. Draw the curve of y = 2 — * — x*. Find the maxi- mum value of the function and the corresponding value of x. • CHAPTER XIV QUADRATIC EQUATIONS 120. Algebraical solution. Plotting the graph of a quadratic function led logically to the solution of a quadratic equation — i.e., an equation of the second degree. The solution by this method is useful and illuminating, but as a method of solving a quadratic equation it is cumbersome, and the accuracy obtainable is limited. We must therefore find an algebraical solution which is certain, universally applicable and capable of any required degree of accuracy. A quadratic equation was solved for the first time when, in § 109, from the curve of y = x % it was found that if x 1 = 3, the corresponding values of x were + 1-73 and — 1-73. The reasoning may be stated thus: ** = 3. *= ± V3 x = + 1-73 or - 1-73. and This is the simplest form of a quadratic equation. It involves the operation of finding a square root ; hence the term " root " as applied to the solution of an equation. In § 115 an important step forward was made. The points where the curve of (x — l) 2 — 4 cuts the * axis, and the function was therefore equal to zero, were found, and the corresponding values of * noted — -viz., 3 and — 1. This means that for these values of x, the expression (x — l) 2 — 4 = 0. They are therefore the roots of this equation. Let the equation be written in the form: (*-l) 2 = 4 .... (A) Then algebraically it is of the same form as the equation F (ALG.) l6 5 i66 TEACH YOURSELF ALGEBRA above, x* = 3. We can proceed with the algebraical solu- tion on the same lines. Taking the square roots of each side: whence x - 1 = ± 2. x — 1 = + 2 or x — 1 = — 2. x = 3 or x = — I. The form marked (A) is the form to which, ultimately, all quadratic equations are reduced; our object always is to reach this form. 121. The method of solution of any quadratic. It was shown that the expression (x — I)* — 4 simplified to x* — 2x — 3. /. the equation which was solved could have been written in the form: x*-2x-3 = . . . . (1) If we start with this equation and wish to solve it, we need to get back to the form: {x — I)' — 4 = . . . . (2) This is the converse operation of changing from (2) to (1), i.e. of obtaining the complete square (2) when we are given (1). Two preliminary steps are necessary. (1) Remove the constant to the right side, as it does not help in finding the square. (2) Divide throughout by the coefficient of**, if this is not unity. Finally we arrive at the form: x* - 2x = 3. It is now necessary to add to the left side such a number ae will produce a complete square. Remembering the work in § 92, we get the following rule : Add to each side the square of half the coefficient of x. QUADRATIC EQUATIONS 167 Then the above becomes x* - 2x + (1)» = 3 + 1. (x - 1)» = 4. and we proceed as before. We will now apply the above method to solve the equation : x i - 6 X + 5 = o. Then x* — 6x = — 5. Add to each side ($) 2 — i.e., 3*. Then x» — 6* + (3) a = — 5 + 9. (*-3)* = 4. Thus we reach the desired form. Proceeding as before: x - 3 = ± 2. x = 3 ± 2. Then x — 3 + 2 = 6 or x = 3 - 2 = 1. .". the solution is x = 5 or x = I. 122. Solution of 2x s -f 5x — 3 = 0. Applying the preliminary steps (1) and (2) of the previous paragraph, we get in succession : and 2x l + 5x = 3 x* + \x = ?. Half the coefficient of x is J, Adding (J) s we get: **+!*+ (5) a = I + ($)«■ Taking square roots of both sides: x + i = ± I and * = _$ + $ = $ or * = -3-2=-¥. x = J or x = — 3. i68 TEACH YOURSELF ALGEBRA 123. Worked examples. Example I. Solve the equation * 2 — x — 1 = 0. Transposing * a - * = 1. Adding (£)* x'-x+ (£)* - 1 + i or and (* - h)* = i * = i±-2- or * = i- x 2 2 Vo 1 - 2-236 = - 0-618. x = I -6 1 8 or x = — 0-6 1 8 (both approx.). Example 2. Solve the equation Zx* — 5x -j- 1 = 0. Applying preliminary steps, § 121: 3** — 5x = — 1 x* - f* = - i Adding ($)»: ' It/ x* _|* + « I* - »' — " J + ft- Taking square roots is 5T- *— S = _ ±3-606 (ap prox.) * = 6 5 ± 3-606 5 + 3-606 8-606 .... •- — 6— = T— 1484 or * = 5 — 3-606 1-394 0-232. 6 6 .-. solution is x = 1-434 or x = 0-232 (approx.). Example 3. Solve the equation: 1 1 I x — 1 x + 2 16* QUADRATIC EQUATIONS 169 First we clear fractions by multiplying throughout by the least common denominator — viz., 16(* — l)(.v 2). Then 16(* + 2) - 16(* - 1) = ( x - \)(x - 2). 16* + 32 — 16* + 16 = x * + x- 2. Adding and transposing: — ** — *=— 48 — 2 or ** + * = 50. Adding (*)*: x* + x+ (*)■ = 50 + I = i|i. ± 14-177 (approx.). 2 x=-\± (* + i)'=±v|?J and or *=-! + * = -*_ 2 14-177 2 14-177 13177 15177 2 2 /. the solution is x = 6-588 or x = — 7-588. Exercise 35. Solve the following equations : = 6-588. = - 7-588. 1. 3* 2 = 12. 2. 4* s - 1 = 0. 3. * 2 -9 16 ~ 9 - 4. (* + 1)2 - 4 = 0. 5. {* - 3)* - 25 = 0. 6. (* + 5)* = 36. 7. (* + J)' = 1. ifei*-* 9. 11. 13. 15. x 2 — 10* + 16 = 0. X s — 2* — 15 = 0. *(* - 4) = 32. 2** — 3* — 5 = 0. 10. ** + * — 12 = 0. 12. * 2 + 3* — 28 = 0. 14. 2* 2 — 7* + 6 = 0. 16. 3* 2 = 7* + 9. 17. 3** + 1 = 5*. 3* -^ =14. * *f +,-£*-* 19. 20. * — 2 = -. * 21. *-9 x+5 3 * * 00 1 l.l ' *— 1 x + 2 16 ijto TEACH YOURSELF ALGEBRA 124. Solution of quadratic equations by factorisation. There is another method of solving quadratics; it is as follows: We first note that if n be any finite number »xo = o, i.e., the product of any finite number and zero is always zero. Conversely, if the product of two factors is zero, then either of the factors may be zero. For example, if a X 6 = then this is true if either a — 0, or b = 0. Similarly if (x-l)(*-3)=0 ... (A) it follows from the above that this equation is satisfied if either x— 1=0 or x — 3 = 0. But if x — 1 = 0, then x =1, and if x — 3 = 0, then * = 3. /. equation (A) must be satisfied if either x — 1 or * = 3. .". 1, and 3, are the roots of the equation: (*_!)(*_ 3) =0, i.e., x % — 4x + 3 = 0. If it is required to solve the equation x* — 4x -f 3 = 0, this can be done by reversing the above steps ; consequently we find the factors of x* — 4* + 3 and so we get : (%-l)(*-3)=0. The method is an easy one, if it is possible to obtain factors. In equations which arise out of practical work this can seldom be done. It is a valuable method in other ways, and can be em- Eloyed in equations of higher degree. If, for example, we now that (*-l)(*-3)(*-4)=0. Then by the above reasoning this equation is satisfied by: x — 1 = 0, whence x = 1 x — 3 = 0, „ % = 3 x — 4 = 0, „ x = 4. QUADRATIC EQUATIONS i 7 . The product of the three factors is an expression of the third degree, since the term of highest degree will be x*. The equation is therefore of the third degree, or a cubic equation. 125. Worked examples. Example I. Solve the equation x* — 2x = 15. Transposing x* — 2x — 15 = 0. Factorising (*-5)(* + 3) = 0. * — 5 = and x = 5 or x + 3 = and x = — 3. .'. the solution isx = 5orx = — 3. Note. — The student should remember that the reasons for this method require that the right-hand side should be zero. Thus it cannot be used in such a case as (*-6)(*-2) = 4. Example 2. Solve the equation Qx(x + 1) = 4. Simplifying and transposing 9x* + 9* — 4 = 0. Factorising (3* + 4)(3*-l)=0. 3* + 4 = and x = — J Zx — 1 = and * = J. .*. the solution is x = — | or x = $. Exercise 36. Solve the following equations by the method of factors: or t - 3) = 0. — 2>» = 0. 1. 3. {x 6. (* + 4)(*-l)=0. 7. 4(3*-7)(2x+ 11] | = 0. 8. ** — 9* + 20 = 0. 10. x* + 2x = 35. 12. x(x - 4) = * + 66. 2. x(x + 5) = 0. 4. (*-l)(*-2)=0. 6. 2(%-3)(* + 7)=0. 9. x* + x = 6. 11. x{x+ 13) +30 = 0. 13. (*-8)(* + 4) = 13. If! TEACH YOURSELF ALGEBRA 14. 16. 18. 19. 20. 21. 22. 23. 2* s - 3* - 5 = 0. 8x % — 14* - 15 = 0. 2x 2 - 11* + 12 = 0. 15. 3*» - ix + 1 = 0. 17. 24** + 10* - 4 = 0. (* — 2)(*-4)(*-5) =0. 2x- l)(* + 3)(* + 2) =0. 'x- !)(*«- 2* — 8) =0. (2x - 5 (** - 5* - 50) =0. (* - a)[x -b) = 0. 24. (2* — c)(x + d) = 0. 126. General formula for the solution of a quadratic equation. We have seen that by simplification and transposing every quadratic can be written in a form such as: 2** + Ix - 4 = 0. in which there are three constants — viz.: (1 the coefficient of x % , 2 ., „ *, 3 the term independent of *. Consequently if we want to write down a general form for any quadratic, letters such as a, b, c can be chosen to represent these constants, so that we could write the general form as: ax 2 + bx + c = 0. The equation above is a special case of this in which a = 2, b = 7, c = - 4. If we solved the general quadratic, ax* + bx + c = 0, the roots would be in terms of a, b, c. We should then have a formula such that, by substituting the values of a, b and c in any special case, we should be able to write down the roots, and no actual solving would be necessary. 127. Solution of the quadratic equation ox* + bx + c = 0. We must use the method of " completing the square " as in §121: ax* + bx + c = 0. Using preliminary step (1) ax 1 + bx = — c. „ (2) **- ( -5*=-?. "73 QUADRATIC EQUATIONS Adding " the square of half the coefficient of * " .J . (b\* b* c X + a X + \2a)=&--a (x + s-J = — ^-j — (adding the fractions). Taking square roots *+S5-± V6 S - 4ac X = -2-« ± 2a Vb 2 - iac 2a or or in full or _ — b ± Vb z — 3oc ~ 2a ~ — b+jy/b* — 4ac X ~~ 2a _ _ — 6 — Vb 2 — iac X ~~ 2a This formula should be carefully learnt : by using it. there is no real necessity for working out any quadratic equation. Before applying it, however, the equation to be solved must be written down with all the terms on the left side. From this formula, since there are always two square roots of a number, Vb 2 — iac must always have two values. Therefore * must have two values and every quadratic equation must have two roots. This fact was obvious in the solution by means of a graph. 128. Worked examples. Example I. Solve the equation 5x* 4- fix = 2. Writing the equation in general form we get: 5** + 9* — 2 = 0. Using the formula -b ± Vb* — iac x = 2a we have for this example a = 5. 6 = 9, c = - 2. 174 TEACH YOURSELF ALGEBRA Substituting y = - 9 ± V9* - (4 x 5 x - 2) 2x5 _ _ 9 ± V81 + 40 10 = - 9 ± VWl 10 -9±11 x = or 10 - 9 + 11 _ 2 10 10 _ — 9 - 11 _ - 20 * _ 10 ~ "To~- x = 0-2 or - 2. Example 2. Solve the eqxtation ——. + § = — ^-j. Clearing fractions by multiplying throughout by 3(*-l)(*-3) 3(x - 3) + 2(x - 1)(* - 3) = 6(2 - 1) 3* — 9 + 2(x» — 42 + 3) = 62 — 6 32 — 9 + 2x* — 82 + 6 = 62 — 6. Collecting like terms 22* — 11* + 3 = 0. Using the formula — b± Vb* — 4ac x = 2a and putting a = 2, 6 = — 11, c = 3, 11 ± V121 — 24 we get 2 = — == — j - li±J^= 11 +9-849 (app.) 4 4 11 + 9-849 20-849 2 = or x — 4 — 4 11 - 9-849 1151 = 5-212 = 0-288. 4 4 .". the solution Is x = 5-212 or x = 0-288. QUADRATIC EQUATIONS 175 Exercise 37. Solve the following equations by using the formula of §160: 1. x* + 32 - 1 = 0. 3. x* -0-42= 1-6. 5. 22* — 62 = 2. 7. 22* + 122 — 7 = 0. 9. 42* + IO2 = 5. 11. 42*= (2 + 4)(2-2). 13. 32* = 72 + 2. 2. 2* - 52 + 2 = 0. 4. 32* — 62 + 1 = 0. 0. 02* = 72 + 3. 8 32* =82 + 8. 10. 42(2 + 2) = 9. 12. 0-9(2 + 1) = 0-8 — *«. 14.*-^+oA-,=t. 15. 0-5 + 2 + 3 11 2 + 7 4 32 + 1 ^ 22 + 1 = 0. 16 ' 2 + 2 2 — 3 '~ »" 17. (2 + 3)(2-5) = 22-2. 129. Problems leading to quadratics. The following examples illustrate the method of solving such problems. Example I. The distance (/») which a body reaches in time t when it is projected vertically upwards with velocity u is given by the formula h = ut- $gl*. If u = 160 and g = 32, find the litne a body takes to rise 240 ft. Substituting the given values in the formula 240 = 160* - 16**, transposing 16/* — 160/ + 240 = 0. /* - 10/ + 15 = 0. — b ± Vb* — 4ac 2a Using the formula on substitution 10 ± VlOO - 60 10± V40 2 2 / = = 5 ± 3-16 approx. t= 1-84 or t = 8- 16. 176 TEACH YOURSELF ALGEBRA The two roots require consideration. Since every quadratic has two roots, when these furnish the answer to a problem the applicability of the roots to the problem must be examined. Sometimes it will be clear that both are not applicable, especially if one is negative. Consider- ing the above problem, when a body is projected vertically upwards its velocity decreases until it reaches its highest point, when it is zero. It then falls vertically and retraces its path. Therefore It will be at a given height twice, once when ascending and again when descending. The value / = 1-84 gives the time to reach 204 ft. when going up, and it is at the same height when descending 8-16 sees, after starting. Example 2. A motorist travels a distance of 84 miles. He finds that if on the return journey he increases his average speed by 4 miles per hour, he will take half an Iwur less. What was his average speed for the first part of the journey and how long did he take for the double journey ? Let * miles per hour be the average speed for the first journey. Then time for the first journey is — hours. Speed for the return journev is (x + 4) miles per hour. 84 .*. time for the second journev is 7. J x + 4 But this is % hour less than the first time. 84 84 T~7+~4 - *- Clearing fractions by multiplying by the L.C.D.— viz., 2x(x + 4). 168(* + 4) - 168* = x(x + 4) 168* -f 672 - 168* = x* + 4*. Collecting and transposing x 1 -f 4* — 672 = 0. Using the formula QUADRATIC EQUATIONS _ - b ± Vb* - 4 ae X ~ ~2i 177 or substituting _4 ± -y/16 - (4 x -672) 2 _ — 4 ± V2704 — 4 ±52 , = _i+JL2 = 24 or * = 2 -4 — 52 = -28. The negative root, although it satisfies the equation above, has no meaning for this problem. .■. average speed for first journey is 24 miles per hour. Time for first journey = g-J = 3| hours. „ second „ = f» = 3 "hours. /. total time = 6* hours. Exercise 38. 1. The sum of a number and its reciprocal is 2-9. Find the number. 2. The area of a rectangle is 135 sq. in. and its perimeter is 48 ins. What are the lengths of its sides ? 3. Solve for = the equation =, — -= = 10. 4. The relation between the joint resistance R and two resistances r, and r 2 is given by the formula I = i + I. If R is 12 ohms and r % is 6 ohms greater than r„ find V, and r t . 6. A formula for finding the strength of a concrete beam is bn 2 + 2am{n — c) = 0. Solve this for n when b = 4 a = 2, c = 8, m = 12-5. 6. The formula giving the sag (D) in a cable of length L and span S is expressed by L = ~ + S. Find S when L = 80, D = 2-5. 7. The product of a number n and 2»» — 5 is equal to 250. What is the value of m? vfi TEACH YOURSELF ALGEBRA «, 8. There is an algebraical formula s = ~{2a -f (n — l)d). If s = 140, a = 7 and <* = 3, find n. 9. By a well-known geometrical theorem the square on the diagonal of a rectangle is equal to the sum of the squares on the two sides. The diagonal of a particular rectangle is 25 in. long, and one side of the rectangle is 5 in. longer than the other. Find the sides of the rectangle. 10. The cost of a square carpet is £24-80 and the cost of another square carpet whose side is 3 ft. longer than that of the first is £38-75. If the cost per square yard is the same for both carpets, find the area of each. 11. There is a number such that when it is increased by 3 and the sum is squared, the result is equal to 12 times the number increased by 16. What is the number? 12. Two adjacent sides of a rectangle are represented by (x + 4) and (x -f 6). The area of the rectangle is equal to twice the area of the square whose side is x. What is the value of xl 13. Find the area of a rectangular plot of ground whose perimeter is 42 yards and whose diagonal is 15 yards. 14. The cost of boring a well is given by the formula, C = 2x + gjr x*. where C is the cost in pounds and x is the depth in feet. If a well cost £280 to bore, how deep was it ? 15. One number exceeds another by 4. The sum of their squares is 208. What are the numbers ? 16. The formula for the sum of the first « whole numbers is iw(n -j- 1). If the sum is 78, how many numbers are there ? 130. Simultaneous equations of the second degree. These are equations which involve two unknowns and which include terms of the second degree. They are also called simultaneous quadratics. The degree of a term is shown either by its index, or, if it contains two or more letters, by the sum of their indices. * 8 . xy, y % are terms of the second degree involving two unknowns, and any of these, together with terms of the QUADRATIC EQUATIONS 179 first degree, may occur in the equations to be solved. Numerical coefficients do not affect the " degree " of a term. It is seldom possible to solve simultaneous quadratics unless they conform to certain types. In this book we shall consider two of these types. 131. First type. When one of the equations is of the first degree. Example. x+y=\ . . . . (1) 3 x i _ xy + y t = 37 m m , . ( 2 ) It is always possible to solve simultaneous quadratics if we can find one letter in terms of the other, the relation being a linear one. In the equations above, from (1) we get: y = 1 — x. This can now be substituted in equation (2). Thus we get: 3x* — x(l — x) + (1 — x)* = 37. In this way a quadratic, with one unknown, is reached, and this can always be solved by the methods previously given. Simplifying 3x* — x + ** + 1 — 2x + x* = 37. 5** - Zx - 36 = 0. This could be solved by factorisation, or using the formula: On substituting x = — b±Vb» — 4ac 2a *r + 3 ± V9 + 720 10 or _ 3 ±27 10 ' 30 24 * = io or -io x = 3 or — 2-4. i8o TEACH YOURSELF ALGEBRA To find y substitute in y = 1 — x. (1) Whenx = 3, y = 1 - 3 = - 2. (2) When x = - 2-4, y = 1 — (- 2-4) = 3-4. /. the solutions are: (1) x = 3. y = - 2- (2) x = - 2-4. y = 3-4. The solutions should be arranged in corresponding pairs. Exercise 39. Solve the following equations: 1. x — y = 2. x l + xy = 60. 3. 2x + y = 5. 5x* - 3xy m> 14. 5. x +y + \ =0. 3x a - 5y* — 7 = 0. 7. 2* 8 — lbs + 4xy = 60. 3x — y = 9. 9. 3x+y = 25. xy = 28. x y * 3* — y = 2. 2. x + y = 7. 3** + w - v* = 81. 4. 3* + y = 8. x* — bxy 4- 8y = 36. 6. 2x + 3y = 14. 4*« + 2xy + 3y a = 60. 8. 2x + 3y = 5. y + 3 *"** 10. 2* - y = 8. 9 + 16= 5 - 132. Second type. Symmetric equations. Example I. An example of this type ts x +y= 19 xy = 84 (2) These two equations are called " symmetric " because, if the letters x and y are interchanged throughout, the QUADRATIC EQUATIONS ,8i equations are unaltered. Other equations which are not strictly symmetric can be solved by the same method as these, and so are included. In these particular equations number (1) is of the first degree, and so the previous method can be used; but this is an easy example to illustrate the special solution which can be employed with this type. The aim of the method is to find the value of x —y. Then, since the value of x + y is known, the rest of the solution is easy. We are given *4-.y = 19 (1) «y = 84 (2) Squaring both sides of (1) x l + 2xy + y* = 361 ... (3) From (2) w 4*y = 336 (4) Subtracting (4) from (3) we get x* - 2xy +y* = 25. x - y = ± 5. From (1) x + y = 19. Adding 2x = 19 ± 5 = 24 or 14. x = 12 or 7. Subtracting 2y = 19 — (± 5) = 14 or 24. Y = 7 or 12. Or using xy = 84 (1) when x = 12. y = 7, (2) „ x = 7. y = 12. The symmetry of the two equations appears in the solution. Example 2. Solve the equations: x*+y* = 89 (1) xy = 40 (2) Both equations are of the second degree. The aim in this case is to obtain both x + y and x —y. From (2) 2xy = 80 (3) 182 TEACH YOURSELF ALGEBRA Adding (1) and (3) x l + 2xy + y % = 169. x+y=±13. Subtracting (3) from (1) x 2 — 2xy + y 2 = 9. *-y = ±3. Associating the equations: x+y=±13. x-y=±Z. Adding 2x = ± 13 ± 3 = ± 16 or ± 10. * = ± 8 or ± 5. To get corresponding values of * we may now substitute in xy = 40. .-. when x = + 8, — 8, + 5, - 5 y = + 5, — 5, + 8, - 8. .'. associating these the solution is: x = + 8, y = + S. x = — 8, y = — 5. x = + 5, Y = + 8. x = - 5. y = - 8. Note. — In the above examples the sum and difference of two numbers, x + y,x — y, are used. The same method would be used if we were given such an example as x + 3y. We should then aim at getting (x + 3y) a , (x — 3y^*, and ultimately x — Zy. There are many other variations in this type of solution. In some, such as ^+y* = a (1) x+y = b (2 xy is not given, but it can be obtained by squaring equation (2) and subtracting equation (1). Exercise 40. Solve the following equations: I. x + y = 8. 2. * +y = 9. QUADRATIC EQUATIONS 3. x —y = 5. 4. ** + y 2 = 17. xy = 24. x+y = 5. 5. x — 2y = 2. 6. x 2 + y 2 = 17. xy = 12. *y = 4. 7. x 2 — xy + y» = 67. 8. x 2 — xy +y 2 xy = 18. x+y = 8. D. x 2 + y* = 45. 10. 2* + 3y = 9. ** — xy + y 2 = 27. xy = Z. 183 xy = 15. 3xy — 42 = 0. INDICES '85 CHAPTER XV INDICES 133. The meaning of an Index. In § 18 it was shown that the product of a number of equal factors such as a x a x a X a can be written in the form a*, in which 4 is an Index which indicates the number of factors. Generalising this, if there be n such equal factors, where n is a positive integer, then a" will be denned as follows: a n = axaxax ...ton factors and a" Is called the nth power of a. 134. Laws of Indices. It was also shown in §§ 19, 20, 21 that when operations such as multiplication and division of powers of a number are performed, the laws which govern these operations can be deduced from the definitions of a power and an index given in § 133. General proofs of these laws will now be given. I. First law of indices: the law of multiplication. The special cases given in § 19 lead to the general law as follows : Let m and n be any positive integers. By definition a"> = axaxax ...tow factors and a n =axaxax . . . to n factor. Then a m x a" = (a X a X a x ... to m factors) x (a x a x a X ... to « factors). But when two groups of factors are multiplied the factors in the groups are associated as one group of factors to give the product (see § 19). 184 A a™ X a" = a X a X a x a x . . . to [m + n) factors = a m * " (by definition). .*. the first law of indices is a m x a" = m + o The law is clearly true when the product involves more than two powers. Thus a m x a" X ar = a m ' " + p. II. Second law of indices: the law of division. To find the value of o m 4- a". With the same definition as before, and proceeding as in the special cases of § 21. a x a X a X ... to m factors a m — a n = a x a X a X to « factors " After cancelling the n factors of the denominator with n corresponding factors in the numerator, there are left in the numerator m — n factors. • m Q m -1. a n = Q m - n Note. — This proof assumes that m is greater than n. If m is less than m, there are n — in factors left in the de- nominator. /. if « > m a m ~- a" = - _„ . This case will be examined later. III. Third law of indices: the law of powers. By the definition of § 133 such an expression as (a 4 )' means the third power of 0*, or (a 1 ) 8 =a* x a* x a* = a t * 1 * 4 = a* (by first index law) In general, if m and n are any positive integers by definition (a m )" = a m x a™ X a m . . . to « factors = a m * m * m • ■• ton terms (first index law) j86 TEACH YOURSELF ALGEBRA .•. the law of powers is (a m ) n = o mxn . For powers of a product such as (<?&)", see § 20. In a similar way it may be shown that (ab) n = o n X b". Exercise 41. Revision Exercises in Indices. Write down the values of the following: 1. 2a* X 3a 6 . 3. K X Ja 3 . 6. 4 X 4 3 X 4 s . 7. x*y 8 * X *V**- 9. a" + " X a™ ~ ". 11. a" + * X a"" 3 . 29. 2 8 ) 3 . 31. 33. 35 x 6 ) 8 . • (f )• 37. («*)_». 39. VaV 41. V* 19 . 2. a* x a*. 4. 2 X 2 2 X 2 3 x 2*. 6. 3a s 6* x 2a 8 & 8 . 8. x"* 1 X X"*- 1 . 10. a"&? X af*i X ic"'. 12. a"> X a" X a 4 . 13. a 8 -f- a 8 . 14. 5x« -T- lOx 3 . 15. 36a 10 -4- 12a 8 . 16. 2» -r 2«. 17. 3 8 -r 3*. 18. (- *') 4- (- x 3 ) 19. 15a«& 3 -r - 3a&. 20. a 2 '' 4- aP. 21. 12x*ry*i -^ 3xPyft. 22. al ° * * a* 23.£x§. 24. a 2 " -5»<*- 1 . 25 fl ni + n -^ (jm - ». 26. a» + « 4 a"- 3 . „ 4a 3 . 6a 3 6 a 30. (3 3 ) 8 . 32. (x 3 ) 8 . 34. (2a*6 3 ) 3 36. (a?) 3 . 38. (3a**) 3 . 40. Vx 8 . 42. VSo*. 43. vW. INDICES 44 187 «• v 9a 8 4~P* 46. Vx*. 135. Extension of the meaning of an Index. It has so far been assumed that all indices are positive integers. The definition of a" — viz. : a n = axaxax . . . to n factors is unintelligible except upon the assumption that n is a positive integer. But Algebra generalises, and we must therefore consider the possibility of attaching a meaning to an index in all cases. 136. Graph of 2*. As a first step let us choose a suitable number, say 2, and plot some of the powers of it — i.e., draw the graph of 2*, in which x represents any index. Calculating the values of these powers for some of the smaller integral values of x, we get a table of values as follows: X 1 2 3 4 y — 2* 2 4 8 16 When these points are plotted they appear to be on a smooth curve, as drawn in Fig. 40. If we are justified in assuming that this curve is con- tinuous and such that all the points on it satisfy the equation y = 2 1 , then it follows that if any point be taken on the curve between the plotted points, its co-ordinates must also satisfy the law. Therefore if any point, A , be taken on the curve at which x = 1-5 and y = 2-8, then it follows that 2 1 * = 2-8. Again at B, where x = 3-2 and y = 9, 2" = 9. Hence, if the assumption is correct, that the curve is a continuous one, and the co-ordinates of any point on it satisfy y = 2*, then it may be concluded that any number 188 TEACH YOURSELF ALGEBRA within the limits plotted can be expressed as a power of 2, and conversely any number can be used as the Index of some power of 2. Fig. 40. A similar curve could be drawn exhibiting the powers of any other number — say, 10 — i.e., we could draw the curve of y= 10'. Thus we could express any number as a power of 10. Reverting to the curve PQ (Fig. 40), since it was plotted by taking values of x from unity, the curve began at P. If it be produced towards the y axis in the way it INDICES ■ 8g seems to curve, it will apparently cut the y axis at the point where y = 1. This suggests that: (1) The value of 2 Z when * = is I— i.e., 2° = 1. (2) The portion of the curve thus drawn should show values of 2 Z between x = and x = 1. For example, when x = J, 2 X is approximately 1-4 — i.e., 2* = 1-4 approximately. Again, the curve evidently does not end at the y axis, but can be further produced. This portion will correspond to negative values of x. Hence we infer that values of 2 1 can be found when the index is negative. 137. Algebraical consideration of the extension of the meaning of indices. From the graph of powers it may be inferred that powers of a number exist whether the index be integral or frac- tional, positive or negative. We must now consider how such indices can be interpreted algebraically. In seeking to find meanings for the new forms of indices we must be guided by one fundamental principle — viz. : Every Index must obey the laws of indices as discovered for positive Integral numbers. Otherwise it cannot be considered as an index. 138. Fractional indices. We will begin with the simple case of a i . To find a meaning for o*. Since «' must conform to the laws of indices ,*, by the first law a* X a* = a* + » = a 1 or a. :. a* must be such a number that on being multiplied by itself the product is a. But such a number is denned arithmetically as the square root of a. o* must be defined as Va. As an example 2* = V2 = 1-414 approx. This agrees with the value of 1-4 found from the graph igo TEACH YOURSELF ALGEBRA (Fig. 40). The reasoning above clearly holds in all such cases, and so we may infer that in general, if n be any positive integer a'n = n/O. To find a meaning for a*. Applying the first law of indices : a» x a* X «* = ««*•+* = a*. a* = &?. Applying the reason generally, we can deduce that if m and n are any positive integers Thus O" = "Jo™. a« = -vT* a = v?. Indices which are in decimal form can be changed to vulgar fractions. Thus: a ' 85 = a* At = Va. 139. To find a meaning for o°. Using the second law of indices, and n being any number, But a™ -7- a™ a" -7- a" = 1. 0° = I. This confirms the conclusion reached in § 136 from the graph. It should be noted that a represents any finite number. Therefore a = 1, whatever the value of a. Graphically, if curves of a 1 are drawn for various values of a, as that of 2 1 was drawn in Fig. 40, all these curves will pass through the point on the y axis which is unit distance from 0. 140. Negative indices. In considering the curve of 2* in Fig. 40, the conclusion INDICES 191 was reached that the curve could be continued for negative values of x. We must now find what meaning can be given algebraically to a negative index. Consider a" 1 . This must obey the laws of indices. .'. by the first law X a +1 = a~ 1+1 = a° Dividing by a +1 we get = 1. 1 a' i.e., by a -1 we must mean " the reciprocal of a ". With the same reasoning and in general a~'= -s a" 3 = -„ a~" X a*" = a _n * n = a° = 1. a _n = — . a" a~" Is defined as the reciprocal of a". The following examples should be noted: 2 . . 2a -8 = -j (negative index applies to a only) Va and generally = a = a". It should be noted as a working rule that when a power of a number is transferred from the numerator of a fraction to the denominator, or vice-versa, the sign of the Index Is changed. 192 TEACH YOURSELF ALGEBRA If « is a positive number a n increases as n increases. .". a~ n or — decreases as n increases, a" Consequently the curve of 2 1 will approach closer to the * axis for negative values of x. This agrees with the course which the curve in Fig. 40 appeared to be taking. 141. Standard forms of numbers. Indices may be usefully employed in what are called standard forms, by means of which we can express clearly and concisely certain numbers which are very large or very small. The method of formation of them will be seen from the following examples: 26 = 2-6 x 10 l 260 = 2-6 X 10 a 2,600 = 2-6 X 10 s 26,000 = 2-6 x 10 1 260,000 = 2-6 x 10 B and so on. When a number is written in standard form, one digit only is retained in the whole number part, and this is multiplied by a power of 10 to make it equal to the number. It should be noted that the Index of the power of 10 Is equal to the number of figures which follow the figure retained in the integral part of the number. Thus in 547,000,000, when written in standard form, 5 only is retained in the whole number part and 8 figures follow it. 547,000,000 = 5-47 X 10 B . Numbers which are less than unity may be similarly changed into standard form by using negative indices. Thus 0-26 = 2-6 4- 10 or 2-6 X 10- 1 0-026 = 2-6 -f- 10 a or 2-6 x 10- a 00026 = 2-6 + 10 8 or 2-6 X 10" 8 000026 = 2-6 -h 10* or 2-6 X 10-«, etc. It should be noted that the numerical part of the Index Is one more than the number of zeros following the decimal point in the original number. INDICES 193 142. Operations with standard forms. In these operations the rules of indices must be observed. Examples. (1) (1-2 x 10 4 ) X (2-3 X 10 8 ) = (1-2 x 2-3) x (10* X 10 8 ) = 2-76 x 10'. (2) (4-8 x 10 8 ) -^ (1-6 X 10- 3 ) = (4-8 -^ 1-6) x (10 8 -H 10- 3 ) = 3 X 10". Exercise 42. Note. — When necessary in the following examples take V2 = 1-414, V3 = 1-732, VB" = 2-236, VlO = 3-162. 1. State what meanings can be given to the following: (1) 3». (2) 81. (3) a*. (4) 2°- 5 . (5) 3 " 25 . (6) a* 75 . 2. State what meanings can be given to the following: (1) 10*. (2) 4*. (3) ol. (4) 10". (5) 2«. (6) 10 125 . 3. Write down as simply as possible the meaning of: (1) 4«. (2) 8«. (3) «"■. (4) ai. (5) a M . (6) a" 4. Find the numerical values of: 1) 4 1S . (2) 16 125 . (3) 9 15 . |4) 100J. (5) 10*. (6) (J)°- s . 6. Find the values of fr*,9*, 3°, 3 1 , 3 1S , 3 a , 3". Plot these and draw the curve which contains the points. 6. Find the values of: (1) 2 a X 2*. (2) 3 X 3* X 3t. 3) 10* -~ 10«. (4) a* x a*. (5) 2» H- 2«. (6) «** -h a os . 7. Write down the meanings with positive indices of: 4-». 3«- a , 2-*, ij, i,. 1(H. 194 TEACH YOURSELF ALGEBRA 8. Write with positive indices: (1) x- 3 . (2) ar*. (4) (*-*)»• (5) Q)" 1 . 9. Find the values of: (1) 8». (2) 25». (4) (&•»)». (5) (3)iv (6) I .'.. * 2 2* (3) (10»)l. (0) 81*. 10. Find the values of: (1) ii)' 2 - , (4) (36)-**. (2) (S)- 8 - (5) (4)". (3) 16-» B . (6) (J)« 11. Find the value of a' X a 4 X «' when a = 2. 12. Write down the simplest forms of: (1) a* X a*. (3) a 3 4- (- «) 8 . (5) at 2 " 4- ar". (2 (4 (6 10 s X 10-*. - a» -5- (- a) X" -r ar*. 13. Find the values of: (1) a 8 X a . (2) a X 1. (3) a X 0. 14. Write down in " standard form " the answers to the following: (1) (2-2 X 10 5 ) X (1-6 x 10 4 ). (2) (7-1 X 10 3 ) X (2-3 X 10 3 ). (3) (4-62 X 10 s ) -f (2-1 X 10 3 ). 4) (7-4 X 10 8 ) X (5 X 1(H). (5) (1-2 x 10- 3 ) X (2-1 X 10*). CHAPTER XVI LOGARITHMS 143. A system of Indices. In the previous chapter it was seen that by means of the graph of 2 1 it is possible, within the limits of the graph, to express any number as a power of 2. This was confirmed algebraically. For every number marked on the y axis and indicated on the graph, there is a corresponding index which can be read on the x axis. These constitute a system of indices by which numbers can be expressed as powers of a common basic number 2. Similarly, by drawing graphs such as 3*. 5*. 10*, numbers can be expressed as powers of 3, 5, or 10, or any other basic number. Thus in all such cases it is possible to formulate systems of indices which, for any number A, would enable us to determine what power that number is of any other number B, which is called the base of the system. This possibility of expressing any number as a power of any other number, and thus of the formation of a system of indices, as stated above, leads to practical results of great importance. It enables us to carry out, easily and accurately, calculations which would otherwise be laborious or even impossible. The fundamental ideas underlying this can be illustrated by means of a graph of powers similar to that drawn in Fig. 40. For this purpose we will use 10 as the base of the system and draw the graph of y = 10*. As powers of 10 increase rapidly, it will be possible to employ only small values for x, if the curve is to be of any use for our purpose. To obtain those powers we must use the rules for indices which were formulated in the previous chapter. From Arithmetic we know that VlO = 3-16 app., i.e., 10* = 316. '95 I'.ifl TEACH YOURSELF ALGEBRA LOGARITHMS «97 Then 10*" = 10* = (10*)* = (316)* = 1-78 app. (by Arithmetic). lO " 76 = 10* = 10* + » = 10* X 10* = 3-16 x 1-78 = 5-62 app. 10 o-i» = io» = (10*)* = (1-78)* = 1-33 app. •I -2 3 4 5 6-7 8 9 I Fig. 41. In this way a table of values for the curve can be compiled as follows: • 01 25 0-25 0-5 0-75 0-875 1 t 10* 1 1-33 1-78 316 5-«2 7-5 10 The resulting curve is shown in Fig. 41. The following examples illustrate the use that can be made of it in calculations. Example I. Find from the graph the value of 1-8 x 2-6. From the graph 1-8 = lO " 26 2-6 - 10°-" 1-8 X 2-6 = 10** X 10°" = iQfrsa * 0-ia (| irst law of indices) = 10°« 8 = 4-6 from the graph. Example 2. Find V*. From the graph 9 = 100>« ^9 = 9* = (lO™)* = lO - 82 (third law of indices) = 2- 1 from the graph. 144. A system of logarithms. Although interesting as illustrating the principles in- volved, the above has little practical value for purposes of calculation, since we depend upon the readings from a curve which is necessarily limited in size and is not sufficiently accurate. For practical purposes it is obvious that tables of the indices used, calculated to a suitable degree of accuracy, are necessary. Such tables have been compiled and are available for the purpose. For their compilation a more advanced knowledge of mathematics is required than is included in this volume. The tables are constructed with 10 as a suitable base. They give the indices which indicate for all numbers, within the scope of the table, the powers they are of 10. G (alo.) ig8 TEACH YOURSELF ALGEBRA Such a table is called a system of logarithms, and the number 10, with respect to which the logarithms are cal- culated, is called the base of the system. A logarithm to base 10 may be denned as follows : The logarithm of a number to base 10 Is the Index of the power to which 10 must be raised to produce the number. 145. Notation for logarithms. The student may wonder why another, and an unfamiliar term is employed as a name for an index. One reason for this will be seen from the following: Let rt be any positive number. ,, x be its index to base 10. Then « = 10*. This is in reality a formula. If it is required to " change the subject of the formula " (see §51) and express x in terms of the other letters, there is a difficulty in doing this con- cisely. Using words we could write: x = index of power of n to base 10. This is cumbersome, so we employ the word " loga- rithm ",* abbreviated to " log " as follows: x = log 10 « the number indicating the base being inserted as shown. If the base is e, we write * = log, n. In this form x is expressed as a function of n, whereas in the form n = 10*. n is expressed as a function of x. The student must be able to change readily from one form to another. Examples. (1) We saw in § 143 that 56-2 = 10**. In log form this is 1-75 = logj 56-2. (2) 1024 = 2 10 . log 2 1024 = 10. • The choice of the word logarithm can bo explained only by the history of the Word. The student could consult A Short History of Mathematics, by W. W. R. BaU. LOGARITHMS 199 (3) 1000 = 10 3 . log 10 1000 = 3. (4) 81 = 3*. log 3 81 = 4. For ordinary calculations 10 is the most suitable base for a system of logarithms, but in more advanced mathematics a different base is required (see § 153). 146. Characteristic of a logarithm. The Integral or whole number part of a logarithm Is called the characteristic. This can always be determined by inspection when logarithms are calculated to base 10, as will be seen from the following considerations: Since 10° = 1, log 10 1 = 10 1 = 10, log I0 10 = 1 10 2 = 100. log 10 100 =2 10 3 = 1000, log !0 1000 = 3 10* = 10,000, log I0 10,000 = 4 and so on. From these results we see that, for numbers between 1 and 10 the characteristic is 10 „ 100 „ „ „ 1 100 „ 1,000 „ „ „2 1000 „ 10,000 „ „ „ 3 and so on. It is evident that the characteristic is always one less than the number of digits In the whole number part of the number. Thus in log, 3758-7 the characteristic is 3 Iog l0 375-87 log 10 37-587 „2 „ 1. Thus the characteristics may always be determined by inspection, and consequently are not given in the tables. This is one advantage of having 10 for a base. 147. Mantissa of a logarithm. The decimal part of a logarithm Is called the mantissa. In general the mantissa can be calculated to any required number of figures, by the use of higher mathematics. In zoo TEACH YOURSELF ALGEBRA most tables, such as those given in this volume, the mantissa is calculated to four places of decimals approximately. In Chambers' " Book of Tables " they are calculated to seven places of decimals. The mantissa alone is given in the tables, and the follow- ing example will show the reason why: Similarly and log 10 168-3 = 2-2261. 168-3 = 10*" 61 . 168-3 + 10 = 10* M » -f 16-83 = 10 2 " 81 - 1 = 10 1 2! ". log 10 16-83 = 1-2261. logj 1-683 = 0-2261 log, n 1683 = 3-2261. (second law of indices) Thus, if a number is multiplied or divided by a power of 10, the characteristic of the logarithm of the result is changed, but the mantissa remains unaltered. This may be expressed as follows: Numbers having the same set of significant figures have the same mantissa In their logarithms. 148. To read a table of logarithms. With the use of the above rules relating to the character- istic and mantissa of logarithms, the student should have no difficulty in reading a table of logarithms. Below is a portion of such a table, giving the logarithms of numbers between 31 and 35. Ko. Log. 1 3 1 4 5 6 7 8 9 1 J 3 4 B 6 7 8 » SI .33 M M 4911 6061 6186 6316 4923 6065 6198 5328 4942 5079 5211 5340 4955 5092 5224 6363 4969 6105 5237 6366 4988 6119 5260 6378 4997 6132 5263 5391 6011 5145 62711 6403 6024 6169 6289 6416 6038 6172 5302 5128 1 1 1 1 1 t 8 8 8 ■1 4 4 4 6 t 6 6 1 II 8 8 8 8 10 9 8 11 11 10 in 12 11 1! 11 M 1 5441 545S 5465 6478 6490 5602 5514 6527 6539 1 6561 s 1 4 i a 1 1 7 ii 10 11 (1! («) The figures in column 1 in the complete table are the numbers from 1 to 99. The corresponding number in column 2 is the mantissa of the logarithm. As previously stated, the characteristic is not given, but can be written LOGARITHMS 201 down by inspection. Thus log I0 31 = 1-4914, log 10 310 = 2-4914, etc. If the number has a third significant figure, the mantissa will be found in the appropriate column of the next nine columns. Thus log, 31-1 = 1-4928, log 10 31-2 = 1-4942, and so on. If the number has a. fourth significant figure space does not allow us to print the whole of the mantissa. But the next nine columns of what are called " mean differences " give us for every fourth significant figure a number which must be added to the mantissa already found for the first three significant figures. Thus if we want log 10 31-67, the man- tissa for the first three significant figures 316 is 0-4997. For the fourth significant figure 7 we find in the appropriate column of mean differences the number 10. This is added to 0-4997 and so we obtain for the mantissa 5007. log, 31-67 = 1-5007. Anti-logarithms. The student is usually provided with a table of anti- logarithms which contains the numbers corresponding to given _ logarithms. These could be found from a table of logarithms but it is quicker and easier to use the anti- logarithms, which are given at the end of this book. The tables are similar in their use to those for logarithms, but we must remember: (1) That the mantissa of the log only is used in the table. (2) When the significant figures of the number have been obtained, the student must proceed to fix the decimal point in them by using the rules which we have considered for the characteristic. Example. Find the number whose logarithm is 2-3714. First using the mantissa — viz., 0-3714 — we find from the anti-logarithm table that the number corresponding is a;iven as 2352. These are the first four significant figures >f the number required. Since the characteristic is 2, the number must lie between 202 TEACH YOURSELF ALGEBRA 100 and 1000 (see § 146) and therefore it must have 3 significant figures in the integral part. /. the number is 235-2. Note. — As the log tables which will be usually employed by the beginner are all calculated to base 10, the base in further work will be omitted when writing down logarithms. Thus we shall write log 235-2 = 2-3714, the base 10 being understood. Exercise 43. 1. Write down the characteristics of the logarithms of the following numbers: 15, 1500, 31,672, 597, 8, 800,000 51-63, 3874-5, 2-615, 325-4 2. Read from the tables the logarithms of the following numbers : (1) 5, 50, 500, 50,000. (2) 4-7, 470, 47,000. (3) 52-8, 5-28, 528. (4) 947-8, 9-478, 94,780. (5) 5-738, 96-42, 6972. 3. Find, from the tables, the numbers of which the following are the logarithms: 1) 2-65, 4-65, 1-68. 2) 1-943, 3-943, 0-943. 3) 0-6734, 2-6734, 5-6734. 4) 3-4196, 0-7184, 2-0568. 149. Rules for the use of logarithms. In using logarithms for calculations we must be guided by the laws which govern operations with them. Since logarithms are indices, these laws must be the same in principle as those of indices. These laws are given below; formal proofs are omitted. They follow directly from the corresponding index laws. (1) Logarithm of a product. The logarithm of the product of two or more numbers Is equal to the sum of the logarithms of these numbers (see first law of indices). LOGARITHMS zo 3 Thus if p and q be any numbers log (/> X q) = log p 4- log q. (2) Logarithm of a quotient. The logarithm of p divided by q is equal to the logarithm of p diminished by the logarithm of q (see second law of indices). Thus log (p 4- q) = log p — log q. (3) Logarithm of a power. The logarithm of a power of a number is equal to the logarithm of the number multiplied by the index of the power (see third law of indices). Thus log a" = n log a. (4) Logarithm of a root. This is a special case of the above (3). Thus log Va — log a" = - log a. 150. Examples of the use of logarithms. Example I. Find the value 0/57-86 x 4-385. Let x = 57-86 x 4-385. Then log x = log 57-86 4- log 4-385 = 1-7624 4- 0-6420 = 2-4044 = log 253-7. x = 253-7. No. | log. 57-86 1-7624 4-:iS5 0-6420 253-7 i 2-4044 Notes. — (1) The student should remember that the logs in the tables are correct to four significant figures only. Consequently he cannot be sure of four significant figures In the answer. It would be more correct to give the above answer as 254, correct to three significant figures. (2) The student is advised to adopt some systematic way of arranging the actual operations with logarithms. Such a method is shown above. 20, TEACH YOURSELF ALGEBRA Example 2. Find the value of 5-672 X 18-94 Let 1-758 5-672 X 18-94 1-768 .-. log x = log 5-672 + log 18-94 — log 1-758 = 0-7538 + 1-2774 — 0-2450 = 1-7862 = log 61-12. * = 61-12 No. 5-672 18-94 1-758 log. 0-7538 1-2774 2-0312 0-2450 01 12 11-7862 or x = 6 1 • I (to three significant figures). Example 3. Find the fifth root of 721-8. Let x = y72T8 = (721-8)*. Then log * = J log 721-8 (see § 149(4)) = J (2-8584) = 0-5717. x = 3-730. Exercise 44. Use logarithms to find the values of the following : 1. 2. 3. 4. 5. 6. 7, 8. 9. 10. 11. 12. 13. 23. 24. 23-4 x 14-73. 43-97 x 6-284. 987-4 x 1-415. 42-7 x 9-746 x 14-36. 28-63 -^ 11-95. 43-97 4- 6-284. 23-4 -7- 14-73. 927-8 4- 4-165. 94-76 x 4-195 -r- 27-94. 15-36 x 9-47 x 11-48 14. 15. 16. 17. 18. (15-23) 2 X 3-142. (5-98)* H- 16-47. (91-5)» 16-92" X (1-059) 8 4-73 X (8-97)' 5-632 X 21-85 9-478) 3 . '51-47)*. (1-257) 8 . If kt 1 = 78-6 find r when re = If Jrer 3 = 15-5, find r when re 19. 20. 21. 22. 57-7 4798 (56-2) s -f- (9-814) 3 ^1-625* x 4 -738. "/61-5 x 2-73. 3142. = 3-142. LOGARITHMS *>3 151. Logarithms of numbers between and I. In § 146 we gave examples of powers ol 10 when the index is a positive integer. We will now consider cases in which the indices are negative. Thus 10 l = 10 10° = 1 10-» = A = 01 /. log 10 10 = I ••• log 10 l = /. log 10 0-l =-1 ">- a = iJl-0-01 .". log 10 001 = - 2 1(H= IS3 =0001 /. log 10 0001 = - 3 etc. From these results we may deduce that: The logarithms of numbers between and 1 are always negative. We have seen (§ 147) that if a number be divided by 10, we obtain the log of the result by subtracting 1. Thus if log 49-8 =1-6972 log 4-98 = 0-6972 log 0-498 = 0-6972 - 1 log 0-0498 = 0-6972 - 2 log 0-00498 = 0-6972 - 3. From the above, log 0-498 = 0-6972 — 1 = - 0-3028. Now, in the logs of numbers greater than unity, the mantissa remains the same when the numbers are multi- plied or divided by powers of 10 (see § 147), i.e. with the same significant figures we have the same mantissa. It would clearly be a great advantage if we could find a system which would enable us to use this rule for numbers less than unity, and so avoid, for example, having to write log 0-498 as - 0-3028. This can be done by not carrying out the subtraction as shown above, and writing down the characteristic as nega- tive. But to write log 0-498 as 0-6972 — 1 would be awkward. Accordingly we adopt the notation 1-6972, writing the minus sign above the characteristic. Similarly and *o6 TEACH YOURSELF ALGEBRA It is very important to remember that 1-6972 = - 1 + 0-6972. Thus in logarithms written in this way the charactlstlc Is negative and the mantissa is positive. With this notation log 0-0498 = 2-6972 log 000498 = 3-6972 log 0000498 = i-6972 etc. Note. — The student should note that the negative char- acteristic is numerically one more than the number of zeros after the decimal point. Example I. From the tables find the logs of 0-3185, 0-03185 and 0003185. Using the portion of the tables in § 148, we see that the mantissa for 0-3185 will be 0-5031. Also the characteristic is — 1. log 0-3185 = 1-5031. log 0031S5 = 5-5031 log 0003185 = 3-5031. Example 2. Find the number whose log is 3-5416. From the anti-log tables we find that the significant figures of the number whose mantissa is 5416 are 3480. A3 the characteristic is — 3, there will be two zeros after the decimal point. .-. the number is 0-003480 (correct to 4 significant figures). Exercise 45. 1. Write down the logarithms of: (1) 2-798, 0-2798, 0-02798. (2) 4-264, 0-4264, 0-004264. (3) 0-009783, 0-0009783, 0-9783. (4) 0-06451, 0-6451, 0-0006451. 2. Write down the logarithms of: (1) 0-05986. (4) 0-00009275. (2) 0-000473. (5) 0-5673. (3) 0-007963. (6) 0-07986. LOGARITHMS 3. Find the numbers whose logarithms are: (1) 1-3342. (4) 2-6437. (2) 3-8724. (5) 1-7738. (3) 5-4871. (6) 8-3948. 152. Operations with logarithms which are negative. Care is needed in operating with the logarithms of numbers which lie between and 1, since they are negative and, as shown above, are written with the characteristic negative and the mantissa positive. A few examples will show the method of working. Example I. Find the sum of the logarithms : 1-6173, 5-3415, 1-6493, 0-7374. Arranging thus 1-6173 5-3415 1-0493 0-7374 5-3455 The point to be specially remembered is that the 2 which is carried forward from the addition of the mantissse is positive, since they are positive. Consequently the addition of the characteristics becomes — 1— 2— 1+0 + 2=— 2. Example 2. From the logarithm 1-6175 stiblract the log 3-8463. 1-6175 3-S463 1-7712 Here in " borrowing " to subtract the 8 from the 6, the — 1 in the top line becomes — 2, consequently on subtract- ing the characteristics we have _2-(-3) = -2 + 3= + l. 208 TEACH YOURSELF ALGEBRA Example 3. Multiply 2-8763 by 3. 5-8703 3 LOGARITHMS 2ot, i-6289 From the multiplication of the mantissa, 2 is carried forward. But this is positive and as (— 2) x 3 = — 6, the characteristic becomes — 6 + 2 = — 4. Example 4. Multiply 1-8738 by 1-3. In a case of this kind it is better to multiply the char- acteristic and mantissa separately and add the results. Thus 0-8738 x 1-3 = 1-13594 -lxl-3=- 1-3. — 1-3 is wholly negative and so wc change it to 2-7, to make the mnutissa positive. Then the product is the sum of 11 3594 5-7 or 1-8351)4 1-8359 approx. Example 5. Divide 5-3716 by 3. Here the difficulty is that on dividing o by 3 there is a remainder 2 which is negative, and cannot therefore be carried on to the positive mantissa. To get over the difficulty we write: -5= -6+ 1 or the log as — 6 -f 1-3716. Then the division of the — 6 gives us — 2 and the division of the positive part 1-3716 gives 0-4572, which is positive. Thus the complete quotient is 2-4572. The work might be arranged thus: 3)6 + 1-3 716 3 + 0-4572 2-4572 Exercise 46. 1. Add together the following (1) 2-5178 + 1-9438 + (2) 3-2165 -f 3-5189 + I 2. Find the values of: (1) 4-2183 - 5-6257. (2) 0-3987 - 1-5724. 3. Find the values of: (1) 1-8732 x 2. (2) 2-9456 x .!. (3) 1-5782 x 5. 4. Find the values of: 1) 3-9778 x 0-r,.-,. 2) 2-8947 x 0-S-l. !3) 1-6257 x 0-6. 5. Find the values of: (1) 1-4798 -f- 2. (2) 2-5637 -f 5. (3) 5-3178 + 3. logarithms: 6138 - 6 -5283. 3297 -r 2-6475. (3) 1-6472 - 1-9875. (4) 2-1085 - 5-6271. (4) 1-5782 x 1-5. (5) 2-9947 x 0-8. (6) 2-7165 x 2-5. (4) 2-1342 x - 0-4. (ffl 1-3104 x - 1-5. (6) T-2976 x - 0-8. 4) 3-1195 -i- 2. 5) 1-6173 + 1-4. 6) 2-3178 -+ 0-8. 153. Change of base of a system of logarithms. Although logs calculated to base 10 are usually employed for calculations, in more advanced Mathematics, as well as in Engineering, the logs which naturally arise are calculated to a base which is given by the series I+ i + n-2 + + 1.2.3^1.2.3.4 + to infinity. This series is denoted by e, and its value can be calcu- lated to any required degree of accuracy by taking sufficient terms. To 5 places of decimals, e = 2-71828. Logs calculated to this base are called Naperlan logar- ithms, after Lord Napier, who discovered them in 1614, using this base. They are also called Natural logarithms or Hyperbolic logarithms. mo TEACH YOURSELF ALGEBRA The student, possessing only tables of logs to base 10 may require to use the logs of numbers to base e, and must therefore know how to find them. The relations between the logs of numbers to different bases is found as follows: Let n be any number. Let a and b be two bases. Suppose that logs to base b are known, and we require to find them to base a. Let log 6 n = y, .*. n = b». Then log n = log a (&») = vlog„6. (§ 149, rule 3.) log a n = log n x log a b. Thus, knowing the log of a number to a base b, we find its log to base a by multiplying, whatever the number, by log„ b. In the above result let b = io and a = e. Then log, n = log 10 n x log c 10. In this result let n = e then log, e = log, e x log, 10 but log, e = 1. log 10 e x log, 10 = 1. .*. in the rule log, n = log, n x log, 10 we can write log, n = log 10 n x ■ . Thus both logs on the right-hand side are to base 10. Now and log 10 e = 0-4343 , ^ 10 = 0T343= 2 - 302G - Hence to change from base 10 to base e, we may use either of the following: (I) log, n = log l0 n x 2-3026 or (2) log, n = log 10 n + 0-4343. LOGARITHMS zu Example. Find log, 50. Using log, 50 = log 10 50 x 2-3026 we have log, 50 m 1-6990 x 2-3026. Evaluating the right-hand side by use of logs we get log, 50 = 3-913. Summary of the laws relating to logarithms, togetherwith some special points the truth of which will be obvious. (1) The logarithm of the base itself is always unity. (2) The logarithm of 1 is always zero, whatever the base. (3) The logarithms of all numbers less than unity are negative. (4) The logarithm of a number is equal to — (the log of its reciprocal). Thus 1 ! (7) (8) l0g a M= — log -. log„ (xxy) = log„ * + log„ y. log„ (x -7-y) = log. * - log„y. log„ x" = n log fl x. log v^=-log a x. Exercise 47. Miscellaneous Exercises in 1. 15-62 x 0-987. 2. 0-4732 X 0-694. 3. 0-513 X 00298. 4. 75-94 X 00916 x 0-8194. 6. 9-463 4- 15-47. 6. 0-9635 -r 29-74. 7. 27-91 -4- 569-4. 8. 00917 -f- 0-5732. 9. 5-672 x 14-83 4 0-9873. 10. (0-9173) » 11. (0-4967)». 12. •^1-715. 13. 7647-2 -f (3-715) 8 . the Use of Logarithms. 14. J (48-62)*. 16. '•728 142* 1-697) ". 17. 19-72) "". 18. 0-478) s >. 19. 5-684)- 1 **. 20. -5173)-* 4 . 21. -^0-01697. 22. (0 -1478) * -f 0- 6982. 23. 70-8172 -h TO^feBS. 2ia TEACH YOURSELF ALGEBRA 24. 9-74' — 5-66*. (Hint.— This should first be factorized and changed to a product.) 9-32 x 0-761 25. 20 7J 647-3 X 3- 2 142 X 10-78" VT8-2 27. V(3-62) a + (5-47)* + (6-91) a . 28. Find the value of r.r 2 when * = 3-142 and r = 16-89. 29. Find the value of Jw" 3 when * = 3-142 and r = 2-9. 30. If V = pv l * find V when v = 6-032 and p = 29-12. 31. If 3* = 24, find x. 32. If ft" = 1-8575 find ft when n = 18. 33. When I - C find / when g = 32-2, / = 2. * = 3-142. 34. From the formula V -fi |gH5 03L find F when g = 32-2, A = 0-627, 1 = 175, Z> = 0-27. 35. If / = 2-- F-, find g when / = 5-304, t = 2-55 it = 3-142. 36- Without using tables find the values of: (a) log 27 -f log 3. (b) (log 16 - log 2) -^ log 2. 37. If M = PR" find M when P = 200, R = 1-05, n = 20. 38. Find the radius of a sphere whose volume is 500 c. ft. V = $nr*. 39. Find the values of (1) log, 4-6 (2) log, 0-062. 40. The insulating resistance, R, of a wire of length / is given by „ 0-42S . d. ~ X l0 &' d\ T Find / when S = 2000, R = 0-44, d t = 0-3, <f, = 0-16. 41. In a calculation on the dryness of steam the follow- ing formula was used: Find q when L, = 850, L = 1000, T 1 = 780, T = 650, f,-l. CHAPTER XVII RATIO AND PROPORTION 154. Meaning of a ratio. Thkre are two ways of comparing the magnitudes of two numbers: (1) By subtraction. This operation states by how much one number is greater or less than the other. If the numbers are represented by a and b, the com- parison is expressed by a — b. (2) By division. By this means we learn what multiple or what part or parts one number is of the other. The latter is called the ratio of the two numbers, and may be expressed by a -f- b, or t, or in the special form a : b. Of these, the fractional form is best suited for manipulation. 155. Ratio of two quantities. The magnitude of two quantities of the same kind, such as two lengths, weights, sums of money, etc., may be com- pared by means of a ratio. To do this the measures of the two quantities are expressed in terms of the same unit by numbers. The ratio of these two numbers expresses the ratio of the quantities. Thus, the ratio of two distances which are respectively a miles and b miles would be r or a : b. The ratio would be unaltered in value if expressed in other units provided they were the same. Thus the ratio of 3 yards to 2 yards is the same as the ratio of 9 ft. to 6 ft. or 108 in. to 72 in. This is obvious from the consideration that, as a ratio can 213 214 TEACH YOURSELF ALGEBRA be expressed by a fraction, it can be manipulated in the same ways as fractions. a a x m Thus and b' a b x m a -~m b -i- m' A ratio Is always a number, either an integer or a fraction (vulgar or decimal), and is not expressed in terms of any particular unit. 156. Proportion. If four numbers a, b, c, d are so related that the ratios ? and ^ are equal, the numbers are said to be In proportion. It follows from the definition of a ratio that a and 6 must represent the ratio of two quantities of the same kind, while c and d must also represent the ratio of two quantities of the same kind, though not necessarily of the same kind as a and b. Thus a and b might represent the measures of two weights, while c and d „ „ „ costs. Continued proportion. If a series of numbers a, b, c, d . . . is such that «_ b _ c _ b~ c~d~ ■ • ' then these numbers are said to be in continued proportion. Thus in the series of numbers 2, 6, 18, 54 . . . the ratios S 18 f — VS — "6T • • • are all equal. The numbers are in continued proportion. Mean proportional. If a, b, c are numbers such that a b 6 = ? b is called a mean proportional between a and c. Then 6 2 = ac and b = Vac. In this way we can find a mean proportional between any two numbers. RATIO AND PROPORTION 157. Theorems on ratio and proportion. The following theorems should be noted: 215 (1) Let Then (2) Let c d' |x bd=- d x bd. ad = be. a c "a" Then from the meaning of a ratio (3) Let Then (4) Let Then (5) Let Then - = - (alternando). a — £ b~d' ad = be (theorem 1). ad _ be cd ~ cd' o b ,. , , , - = - (tnverlendo). a _ c b~d' a + b ~b~ a y c + d d c d- (componendo). ?->-!-•• a-b c-d d = — -i— (dividendo). (6) By division of (4) by (5) a + b _ c + d a — b~ c — d' This is called componendo and dividendo. I|6 TEACH YOURSELF ALGEBRA 158. An Illustration from Geometry. The following illustration from Geometry is given as being of great importance. For a proof of it and for a geometrical treatment of ratio and proportion the student is referred to any text-book on Geometry. Similar triangles. Triangles which have their corre- sponding angles equal are called similar. In such triangles, the ratios of corresponding sides are equal. In Fig. 42 the triangles ABC, A'B'C are similar — i.e., Kig. 42. LA = LA', LB = LB',LC = LC. Corresponding sides are those which are opposite to equal angles. Denoting the lengths of the sides opposite to these by a, b, c, a', b', c' as shown, then by the theorem stated above a a 6 b' c It follows from theorem (3) of § 157 a a' b V , a a' r = T?. " = — an d - — ~> b b c c c c i.e., the ratios of the pairs of sides containing the equal angles are equal. Each pair of equal ratios gives a set of numbers in proportion. 159. Constant ratios. Let AOB be any angle (Fig. 43). RATIO AND PROPORTION 217 On the arm OA take points P.Q, R . . . and draw PK, QL, RM perpendicular to the other arm. Then the As OPK, OQL, ORM are equiangular and similar. .-. by the theorem of § 158 PK _ QL RM OK OL OM Clearly any number of points, such as P, Q, R, can be taken and the ratio of all pairs of sides such as the above are equal. We can therefore say that for the angle AOB all such ratios are constant in value. This constant ratio is called in Trigonometry the tangent of the angle AOB. It is abbreviated to tan AOB— i.e., PK QL RM OK = OL = OM = tanAOB - A similar result is true for any other angle. .'. every angle lias its own tangent or constant ratio, by which it can bo identified. Referring to §§ 72 and 73 it will be seen that the gradient of a straight line, represented by m, is the tangent of the angle made by the line with the x axis. PO In Fig. 24, for example, the ratio jjk is constant for every point on the line, and is the tangent of the angle POQ. In the general equation y ' = mx 4- b, m represents the tangent of the angle made with the axis of x. 160. Examples of equal ratios. Examples of equal ratios frequently occur in Mathematics, and the following theorem, in different forms, is sometimes useful. 2l8 TEACH YOURSELF ALGEBRA Let f = % = - f be equal ratios (there may be any number it J of them). Let k represent their common value. Then a -i- e -k l~ i~ J~ a = bk c = dk e =fk. These results make possible various manipulations. For example, by addition a + c + e = bk + dk + fk = k(b + d+f). a + c + e = k, b + d+f and is therefore equal to each of the original ratio. This can be varied in many ways. For example, by multiplication of the three equations above: 2a = 2bk 3c = Zdk 7e = Ifk. .". by addition 2a + 3c — 7e = 2bk + Zak — Ifk = k(2b + 3d - If). - 2a t oj~I 4= k = each of the ° ri g' nal rati °- Zo -j~ ott — if Exercise 48. 1. Write dovm the following ratios: (a) £a to b pence. (b) p tons to q tons r lbs. 2. Write down the ratio of " a mile an hour " to " a foot per second ". 3. If the ratio a : b is equal to the ratio 5 : 8, find the numerical values of the following ratios : »i't (2) a 2 : b*. (3) 2a : 36. RATIO AND PROPORTION 2IQ 4. (a) Two numbers are in the ratio of 4:5. If the first is 28, what is the second ? (6) Two numbers are in the ratio of a : b. If the first is x, what is the second ? (c) Two numbers are in the ratio of 1 : *. If the second is a, what is the first ? 5. If the ratio -j — is equal to the ratio -, find x. " T C X 6. A rectangle of area A sq. ft. is divided into two parts in the ratio p : q. Find expressions for their areas. 7. A piece of metal weighing a lbs. is divided into two parts in the ratio x : y. What are the weights of the parts ? 8. If -, r, -, - are four numbers in proportion, find x. 9. What number must be added to each term of the ratio 11 : 15, so that it becomes the ratio 7:8? 10. Find the mean proportional between: (1) ab and be. (2) 8a s and 2b*. (3) a(a + b) and b(a + b). 11. What number added to each of the numbers 2, 5, 8, 13 will give four numbers in proportion ? 12. If | = j, find the value of JL±-J. 13. Find the ratio of T when o (1) 3a = lb. (3) 16a 2 = 256*. «H- 14. (1) What is the result when 270 is altered in the ratio 5:3? (2) What is the result of altering j in the ratio 6:5? (3) What is the result of altering J in the ratio a : b ? 15. If a, x, y, b are in continued proportion, find x and y in terms of a and 6. CHAPTER XVIII VARIATION 161. Direct variation. In Chapter XIII examples are riven of a variable quantity the value of which depended on the value of an- other variable, and is called a function of it. There are very many different forms which this variation may take, and in this chapter we shall examine one of the most important of them. We will begin with a very simple example. If a man is paid at a certain rate for every hour that he works, the total amount he earns over any period depends on the number of hours he works. If he doubles the number of hours he works, his earnings will be doubled. Generally the ratio of the amounts he earns in any two periods is equal to the ratio of the number of hours worked in the periods. If 7\ and 7" 2 represent the number of hours worked in two periods and if W t and W t represent the amounts of wages earned in them, then W,~T. These four quantities are in proportion (see § 108), Hence, when the relations between the two quantities can be expressed in this way we say that The wages are proportional to the time worked or the wages vary directly as the time. The wages are a function of the time, and the words " proportional to " or " vary directly as " arc used to define the exact functional relations which exist between the two quantities. 220 VARIATION 22i Direct variation may be defined as follows: Y Y If two quantities Y and X are so related that v * = -^, where X x and X % are any two values of X and Y t and Y. are corresponding values ofY, then Y is said to be proportional to X, or Y varies directly as X. In order to discover whether or not one quantity varies directly as another, this simple test can be applied. // one quantity is doubled, is the other doubled in conse- quence ? Or, more precisely, if one quantity is altered in a certain ratio, is the other altered in the same ratio ? The sign cc is used to denote direct variation. Thus in the case above we would write Y cc X. 162. Examples of direct variation. (1) The distance travelled by a motor mo\-inp with uniform velocity varies directly as the time. (2) The weight of an amount of water is proportional to the volume. (3) The circumference of a circle varies directly as the diameter. (4) The electrical resistance of a wire varies directly as the length. 163. The constant of variation. If y cc x then y = kx, where k Is a constant. In § 100 the common value of a number of equal ratios was represented by k, a constant number for these ratios. When one quantity varies directly as another, then we have seen that the ratio of corresponding pairs of values of the variables is constant. Consequently, as in § 160, this constant is usually represented by k, and by means of it the relation between the quantities can be expressed as a formula. For instance, in Example (1) of § 162 it was stated that for a body moving with uniform velocity, distance moved varies directly as time. Let s represent any distance travelled. ,, t ,, the corresponding time. Then, since S varies directly as t, any ratio of correspond- ing values of these is constant. 222 TEACH YOURSELF ALGEBRA Let k represent this constant. Then by definition ? = k. s = kt. This is the form of the law connecting s and t, but it cannot be of much use until the value of k is known for this particular case. To find this, we must know a pair of corresponding values of s and t. Thus if we are told that the car moves 40 ft. in 2-5 sees., then, substituting the values for s and t, we get 40 = k X 2-5. k = 40 4- 2-5 = 16. .'. the law connecting s and / for this particular velocity is s= I6t. 164. Graphical representation. If x and y represent two variables such that y oc x, then, as shown in § 163, y = kx. The form of this equation is the same as y = mx, the graph of which was shown in § 72 to be a straight line passing through the origin, m being the gradient of the line. .'. y = kx represents a straight line of gradient equal to k. Consequently the graphical representation of the varia- tion of two quantities where one varies directly as the other is a straight line passing through the origin. 165. To find the law connecting two variables. The engineer and the scientist frequently require to know the law connecting two variables, corresponding values of which have been found by experiments. The law may assume many forms, but in some cases there may be reason to suppose that one of the quantities varies directly as the other — i.e., the law may be of the form y = kx. From the results of the experiment we can proceed to determine: (1) Is the law one of direct variation ? (2) If it is, and the law is of the form y = kx, the value of k must be found. VARIATION *23 To determine (1) the results are plotted. Then: (a) If the graph Is a straight line, the law is one of direct variation. (b) If the straight line passes through the origin, the equation is of the form y = kx. Then k has to be determined. Graphically, k can be found as in § 159 by finding the tangent of the angle made with the axis of x. Algebraically as shown in § 163. A pair of corresponding values of x and y is chosen from the graph. These are substituted in y = kx, and so k is determined. 166. Worked example. A spiral spring is extended by hanging various weights from it. The amounts of extension of the spring for different weights were observed and tabulated as follows : Weight in lbs. 01 0-2 0-3 0-4 0« Extension in ins. . 015 03 044 0-6 0-75 08 07 06 Hi ^ 05 o « 04 v * 02 gst ■WXi \ fiif I t ffl [ra Hi ^::: . u:::^ • •-t- -.y/ 1 .-.:.:. .... i.-.r fm ' -7*- 5 1 ' T mf an mm "<■-■- '• llillhl :::: . .... 01 0-2 03 Values of L Fig 44. 04 0-5 "1 TEACH YOURSELF ALGEBRA VARIATION «3 From these results discover Hie law connecting the attached weight and the extension of the spring. The graph resulting from plotting these results is shown in Fig. 44. It is a straight line passing through the origin, though one of the points, corresponding to a weight of 0-3 lbs., lies slightly off the line. This is to be expected in experimental results Also the line passes through the origin. .-. as shown in § 165, the law connecting the weight and extension is one of direct variation — i.e.. The extension varies directly as the attached load. Let E = extension „ L = load then E oc L and E = kL. To find k a pair of values is taken, at the point P where g = 0-75 and L = 0-5. Substituting these 0-75 = k x 0-5 whence k = 1-5. .'. the law is £ = I -SL 167. y partly constant and partly varying as x. The case frequently occurs, in practical work, of a vari- able quantity which in part varies directly as another quantity and in part is constant. There is an example of this in § 67 in the problem con- cerning the profits of a restaurant. The profit depends on 1) The number of customers which is variable. '2) The overhead charges which are constant. It was found that y = ax — b was the law which con- nected these, where y, the profit, varies directly as x, the number of customers, and also depends on the constant charges b. In general all such cases can be represented by the equation y = kx + b. This is the equation of a straight line, but Ic does not pass through the origin (§73). It contains the two con- stants k and b, which must be determined before the law connecting x and y can be stated. If two pairs of values of * andy are known the solution can be found as shown in § 59, example 2. In practical work pairs of values are louncl by experi- ment. The worked example which follows shows the method of procedure in such cases. 168. Worked example. When two voltmeters are compared they have readings corresponding to C and K below. C .1-9 2-75 3-8 4-8 6-8 K . 1 5-75 8-3 114 14 18-8 It is thought that C and K are connected by a law of the form K = mC + b. Test this by plotting the points and find lite values of m and b. The law A' = mC + b is linear — i.e., it is in the general form of the equation of a straight line. To test them we must find if the plotted points be on a straight line. Comparison of the two sets of values suggests that the ?o IS K io m T ■ FTr "7* tTrf TTTTT ". ;r ' ; i ; t ttttr , — r" : : ; . iffTIl TPT t TTJT H It — ■■*•■■ — ■■■• ■ ■•■ M" ■••* ■ ■•• ■■■■ I ■ ■■•a ..... • ■■•■ ■ ■■■■ ■ ■■■■ *■■■■ ■ ■■■■ >■■■■ >■■ ■■■ ■■■ ■ ■« ■ ■■ ■■■ ■ *■■ ■ ■■■ ■•■■ ■ ■■■ Mm ■•■a ■ ■•• "»■-* ■ ■■a ■ *■■ ■ ■■■ ■ ••■ 33 ■•■■ ■ ••■ ■ •>> ■ ■•■ ■ ■■■ • •■• ■ ••• • ■■■ ■ ■•■ ■ ■■■ -••■ ■ ■■■ ■ ■•■ ■ ■a* ■ ■■■ ■ ■•■ >■■■ • ■■■ ■•a* aaaa ■ •■a ■ ■■■ — ■ •■■ ■ ■•■ ■ ■■• ■ ■■■ ■■■■ ■ ■■■ ■ ••■ ■ ■■■ • ••» ■■■■ ■ •■a ■ ■-> :■:: ■ ■■■ ■ BBS ■ III ■ aaa aaaa ■ ■■• ■ aaa ■ aaa 3 C Pto l">. 226 TEACH YOURSELF ALGEBRA VARIATION (1) (2) scale units for C on the axis of x should be larger than those for K on the y axis. When the points are plotted they are seen to lie approxi- mately on a straight line, slight deviations being due to experimental errors. When this line is drawn as evenly as possible, it appears as in Fig. 45. Two suitable points, A and D, are selected on the line, and their co-ordinates are as follows : For A, C = 3-4, K = 10. „ B, C = 5-3, K = 15-5. These are to satisfy the equation K = mC + b. Substituting we get: 10 = 3-4m + b 15-5 = 5-3m + b Subtracting, l-9w = 5-5 55 on m — r7j = 2-9 approx. Substituting for m in (1), 10 = (3-4 x 2-9) + b. b = 10 - 9-86 = 0-14. .-. the law is K = 2-9C + 014. Exercise 49. 1 . The following are examples in which the value of one quantity depends on another. State in each case whether or no it is a case of direct variation : (a) Distance and time when a man runs a mile race. (b) Interest and time when money bears interest at a fixed rate. (c) The logarithm of a number and the number itself. {d) The y co-ordinate of a point on a straight line and the x co-ordinate. (e) The cost of running a school and the number of scholars. 22 7 Hence find 2. If y = kx and y = 8 when * = 7, find k. y when * = 40. 3. If y is proportional to x and y = 10 when x = 4, find y when x = 15; also find x when y = 8-4. 4. If y oc x, and when y — 16-5, * = 3-5, find the law connecting x and y. Hence find x when y = 21. 5. The distances travelled by a body from rest were as follows: Time (sees.) 1 2 3 4 Distance (ft.) . 3-2 6-4 9-6 12-8 Plot these and find if distance varies directly as time. If it does, find: (1) The law connecting time (t) and distance (s). (2) The distance passed over in 2-8 sees. 6. The extension (E) of a spring varies directly as the force {W) by which it is stretched, A certain spring ex- tended 2-4 in. when stretched by a weight of 4-5 lb. Find : (1) The law which connects them. (2) The extension due to a weight of 7 lb. 7. The law connecting two variables x and y is of the form y = kx + b, where k and b are constants. When x = 10, y = 11, and when x = 18, y = 15. Find * and 6 and state the law. 8. In a certain machine the law connecting the applied force (£) and the load (W) was of the form E = aw + b, where a and b are constants. When w = 20, E = 1-4, and when W = 30, E = 2. Find the law. 9. In experiments to determine the friction (F lb.) between two metallic surfaces when the load is W lb., the following results were found : w 3 5 7 10 12 F 0-62 1-5 2-4 3-6 4-4 228 TEACH YOURSELF ALGEBRA Assuming W and /•" to be connected by a law of the form F = all - b, find the law by drawing the average straight line between the points. Other forms of direct variation 169. y varies as the square of x — i.e.,y oc x*. If the sides of a square be doubled, the area is not doubled, but is four times as great ; if the sides be trebled the area is nine times as great. The area of a square varies directly as the square of the length of Its sides. The ratio of the area of a circle to the square of the length of its radius can be shown experimentally to be constant. If A = the area and r = radius, then -. is constant for all circles. This constant is represented by the special symbol n, and its value is approximately 3-1416. /. the area of a circle varies directly as the square of the length of Its radius. Students of Mechanics will know that the distance passed over by a body moving with uniformly Increasing velocity Is proportional to the square of the time. A special case is a falling body. If s = the distance fallen and t = the time taken in sees., then s oc I 2 and s = kt l . Experiments show that k = \g, where g is an absolute constant whose value is approximately 32-2. The graphical representation of y = kx % is that of a quadratic function (see Chapter XIII). For different values of k, the graph is a parabola, symmetrical about the y axis and having its lowest point at the origin (§ 111). 170. y varies as the cube of x — i.e.,y oc x 3 . If the edge of a cube be doubled, the volume Is 8 times as great. The volume varies directly as the cube of the length of an edge. The volume of a sphere also varies as the cube of the radius. If V = the volume and r = the length of the radius, then V oc r> and V = kr>. VARIATION 2a9 It can be shown that k = f"- V = t*r». The graph of y — x 3 is a curve, as shown in Fig. 46. It is called a cubical parabola. The student should make a table of values and draw the curve. (U ::: m: n -.... . • :::: — (a, ffl > 30 ffHffj| ': ill 'Wm'-i i Hit Fig 46. 171. y varies as Vx or x', i.e., y oc Vx. This form of variation, besides arising in various physical examples, may also be regarded as the inverse of y = x*. Since H alg.i y = kx* . 1 ** = &■ = sl\ xV ~y- 2 3 o TEACH YOURSELF ALGEBRA Since Jrisa constant, then x oc Vy. The graph of y = Vx is a parabola, as in Fig. 47. It is the same curve as Fig. 31, as will be realised from §§ 108 and 109, but being the inverse, it is symmetrical about the x axis. Negative roots will be found on the part of the curve below the x axis. 2 I -I -2 -3 ffl :::::::::::::::: ::::::!: *■■■■■«*■■■ ::::::::::: :;: ===::::: ::::::sv-: ftpjffl *■-■■•■-> If !■■■■■ ■•• ")*•«■• \H :::::::: ^::::r...: ■HJjMiuii "•■■■••■i ■■-• I::!!!:::!: ■ ■■ iv;::::::::: ::::::::::::::::: Fig. 47. 172. Inverse variation: y oc -. Let as and y be two numbers such that their product is constant — i.e., xy = k, then y = - or * = -. J x y Each quantity can be expressed in terms of the reciprocal or inverse of the other. VARIATION *3' Many examples of this occur in Mathematics. The following is a simple case. Let x and y be the sides of a rectangle of area 60 sq. in., then xy = 60. The lengths of the sides may be varied in very many ways, their product always being equal to 60. If x is increased, y will be decreased, and vice versa. If x be doubled, y will be halved. In general, if x be changed in a given ratio, y will be changed m the inverse ratio. Hence we say that y varies Inversely as x — i.e.,y oc -. Hence y = -. J x Among the many examples of inverse variation we may note: (1) Time to travel a given distance varies inversely as the speed. If the speed be doubled, the time is halved. (2) The volume of a fixed mass of gas varies inversely as the pressure on it, the temperature remaining constant. If p = the pressure and v = the volume. I k p oc - or * = -. r v r v (3) The electrical resistance of a wire of given length and material to the passage of a current through it varies inversely as the area of the cross-section oj the wire. If R = the resistance, and A = the area of the cross section, then 173. Graph ofy = -. *«2«*-T In its simplest form, when k = 1. the equation becomes y = -. The graph of this function presents some new n* TEACH YOURSELF ALGEBRA difficulties which will be apparent on drawing the curve. The following is a table of values formed in the usual way: * i i 1 2 3 4 y • < 2 1 i \ i A similar set of values can be tabulated for negative values of x, the corresponding values of y being negative. m f 5 !i; Fig. 48. The curve, which is called a hyperbola. Is shown in Fig. 48. It consists of two branches, alike in shape, one for +ve values and the other for — ve values of x. The follow- ing important features of this curve should be noted. VARIATION *33 (1) As x Increases, j I or - decreases. When x becomes very great, y becomes very small, and the curve approaches very close to the axis of x. (2) As x decreases, y or - Increases. When x is very small, y is very large, and the curve approaches very close to the axis of y. Both these features of the curve are repeated for — ve values of x. It may be noted that the curve is symmetrical (i) about the line through the origin making 45° with the x axis — i.e., the line y = x ; (ii) about a line through the origin at right angles to this — i.e., the line y = — x (see §§ 72 and 73). k For other values of k the curve of y = - is always a hyperbola. 174. Other forms of inverse variation. (1) One quantity may vary inversely as the square of another quantity — i.e., y oc -», whence y = -j. In X X electricity, for example, the force between two magnetic poles varies inversely as the square of the distance between them. Many other physical laws involve this form of variation. (2) Another form of variation is that in which one quantity varies inversely as the cube of another — i.e., y oc ^ and y = ^. Generalising, y may vary directly or inversely as any power of x — i.e., y cc x" ory cc — . In all cases of direct variation the same method, the introduction of the constant k, is followed and the evaluation of k proceeds on the same lines. 175. Worked examples. Example I. If y varies as the cube root of x, and if y = 3 *34 TEACH YOURSELF ALGEBRA when x = 64, find the formula connecting the variables. Hence find x when y = *£. (1) Since y cc x* y = kx*. k = y -r- x*. Substituting the given values y = 3 + 64* = 3 +4 or j. .-. the law is y = |^*. (2) When _y = V. we have y = j-v^. •V^ = ^ -T- } = 5. x = 5»=l25. Example 2. 77«e <i«e of vibration of a simple pendulum varies as the square root of its length. If the length of a pendulum which beats seconds is 39 ins., what will be the lime of vibration if its length is increased by 3 ins. ? Let / = length of pendulum. „ t = time of vibration. Then / oc yT and / = Wl But when / = 39. t = 1. 1 = AV39. 1 k = When / = 42 ins. V39' Using logs, this is found to be 1 -038 approx. t = 1-038 sees. Exercise 50. 1. If y is proportional to x 1 and when x — 15, y = 200, find the equation connecting x andy. Findy when x — 8-5. VARIATION «35 2. If y oc x* and x = 2 when y = 2, find the law con- necting * and y. Then find y when x = 3. 3. If y oc -j, fill up the blanks in the following table: X i 2 3 y t J 4. If v cc V* and if ;y = 3-5 when * = 4, express y in terms of *. What is y when * = 25 ? 5. 11 y cc x 3 and if ^ = 6 when .* = 4, find the value of y when * = 16. Also find x when v = 3. 6. If y is inversely proportional to x, and x = 5 when y = 6, find the law connecting x and y and find x when y = 20. 7. If y cc -, fill up the gaps in the following table: X 1-2 6 y 6 in 0-8 8. The force F, in dynes, which acts between two magnetic poles is inversely proportional to the square of the distance (d) between them. Express this as a formula if F = 120 dynes when d = 4 cms. 9. The square of the distance of the visible horizon varies as the height of the place of observation above sea-level. If this distance is 24 miles when seen from a height of 384 ft., what is the distance seen from a height of 726 ft. ? 10. The weight of a body above the ground, as indicated by a spring balance, is inversely proportional to the square of its distance from the centre of the Earth. If a body weighs 100 lb. on the surface of the Earth, what will it weigh when 100 miles above the surface? (Radius of Earth = 4000 miles (approx.).) 11. The diameter (d) of a shaft is proportional to the cube root of the horse power (H) which it is required to trans- mit. If the diameter necessary to transmit 12 h.p. is 2 3 6 TEACH YOURSELF ALGEBRA 2 ins., find the formula which connects them. What horse power can be transmitted by a shaft of 3 ins. diameter ? 12. For a given source of light the intensity of illumina- tion (/) is inversely proportional to the square of the distance (d). A surface is illuminated with a certain intensity when at a distance of 5 ft. At what distance must the surface be placed so that the intensity of illumination is 1$ times as great? 13. There is a probability that if a man stands at so short a distance as d ft. from the muzzle of a gun which discharges a projectile of weight w lb., his sense of hearing will be hurt. If d is proportional to the sixth root of w and d is 10 ft. for a discharge of a 64-Ib. shot, what is d for the discharge of a 9-lb. shot? 14. If y oc x lA and if y = 354-5 when * = 15, find the law connecting y and x. 176. Functions of more than one variable. Example I. It is proved in elementary Geometry that the area of a triangle is given by the formula A = ibh where A = the area of the triangle b = the length of a base h = the corresponding altitude. Both b and h are variables, and the value of A depends on them both. .\ A is a function of the two variables b and h. In any triangle let the height remain constant, but the base variable, then if the base be doubled, the area will be doubled. If the base remain constant and the height be trebled, the area will be trebled. Now suppose botli base and height to vary ; let the base be doubled and the height trebled, then the area will be 2 x 3 — i.e., 6 times greater. We can infer, then, that if both base and height vary, the area varies as the product of base and height — i.e., A oc b x A. A = kbh. VARIATION *37 From geometrical considerations we know that in this case k = J. Example 2. In § 172 it was stated that the volume of a given mass of gas, at a constant temperature, varies inversely as the pressure on it. It can also be shown by experiment that if the pressure be kept constant and the temperature varied, then the volume varies directly as the absolute temperature. If both temperature and pressure vary, then the volume varies directly as the absolute temperature and Inversely as the pressure. Let v = volume „ T = absolute temperature •> P — pressure then P Example 3. If an electric current passes through a wire, it encounters resistance. This resistance varies as the length of the wire (§ 162), and in a wire of given length— i.e., constant — it varies Inversely as the cross section of the wire (§ 172). .". if R = the resistance / = the length A = the cross section then R varies directly as / and inversely as A. Thus and Roc t A I R = kx^. A 177. Joint variation. The variation of a quantity due to two or more variables is sometimes called joint variation, and the quantity is said to vary jointly as their product. In dealing with problems involving joint variation, the same procedure with regard to the constant of variation and its determination is followed as in the previous case. The following examples will serve to illustrate it. 238 TEACH YOURSELF ALGEBRA 178. Worked example. Example I . A quantity represented by y varies directly as x and inversely as z 3 . // is known that when x = 15, z = 12 and y = S V Find the law connecting the quantities. We are given that x Substituting the given values ' -*x^ Sl , — « A J23 - • the law is * — 5T ~ 123 r- Example 2. TA« /orce between two magnetic poles varies jointly as the product of their strength and inversely as the square of the distance between them. If two poles of strength 8 and 6 units repel one another with a force of 3 dynes when placed 4 cm. apart, with what force will two poles whose strengths are 5 and 9 units repel one another wlien 2 cms. apart? Let F = the force „ mx, m, = the pole strengths „ d = the distance apart then FocSjp. F-*X«ft Substituting the given values 3 = * X 8x6 4- * t. 3X 4» . * ~ d* * 5x9 In the second case F = — r - = 1 1*25 dynes. VARIATION «39 Exercise 51. 1. Express the following statements in the form of equations : la) y varies jointly as x and z. \b) y varies directly as x and inversely as the square of z. (c) y varies directly as the square root of x and inversely as z. (d) The volume (K) of a cylinder varies jointly as the height (h) and the square of the radius of the base (r). (e) The weight (W) which can be carried safely by a beam varies inversely as the length (/), directly as the breadth (b) and directly as the square of the depth (d). (J) y varies directly as the square of x and inversely as the cube root of z. 2. If y varies directly as * and inversely as z, and if y = 10 when x = 8 and z = 5, find the law connecting x, y and x. Also find y when * = 6 and z = 2-5. 3. If y varies jointly as x and z s , and if y = 13J when * = 2-5 and z = 4. nt -d the law connecting the variables. Also find x when z = | and jy = 54. 4. ^ varies directly as x 1 and inversely as Vz. When x = 8 and z = 25, y = 16. Find -/ when x = 5 and z = 9. 5. The load that a beam of given depth can carry is directly proportional to the breadth and inversely pro- portional to the length. If a beam of length 7 ft. and width lj in. can support a load of 4 tons, what load can be sup- ported by a beam of the same material 5 ft. long and 2£ ins. wide ? 6. The resistance of a wire varies directly as its length and inversely as its sectional area. If the resistance of 500 yds. of copper wire of diameter 0028 in. is 19 ohms, find the resistance of 1 mile of similar wire 0-16 in. in dia- meter. 7. The number of heat units (//) generated by an electric current varies directly as the time (/) and the square of the voltage (E) and inversely as the resistance (R). If H = 60, when t = 1, E = 100 and R = 40 find the law connecting them. 240 TEACH YOURSELF ALGEBRA Also find: (1) The value of H when E = 200. R = 120 and t x= 300. (2) The value oi t when E = 120, R = 90 and H - 5760. 8. The load raised by a winding engine varies directly as the steam pressure and inversely as the diameter of the winding drum. If a load of 45 cwt. is raised by a dram of 10 ft. diameter when the steam pressure was 90 lb. per sq. in., what load should be raised by a dram of 12 ft. diameter if the steam pressure is 75 lb. per sq. in. ? CHAPTER XIX THE DETERMINATION OF LAWS 179. Laws which are not linear. In the preceding chapter we considered the determinations of laws which were linear, and which were arrived at by using experimental data. But such laws are not always represented by straight lines. They may involve powers of a variable such as were considered in § 169 and onwards. In these cases, when the results of the experiments are plotted, they will lie on a portion of a curve which might be one of those illustrated m the preceding chapter or of many others. In practice, when only a small portion of the curve can be drawn, it is impossible to identify what curve it is. There are two devices, however, by means of which a straight line can be obtained instead of a curve. The identification can then be made by the methods previously considered. 180. y = ax" -f- b. Plotting against a power of a number. If the law which we require to find is of the form y = ax* -f b or y = ax*, where n Is known, then we plot corresponding values of y and x" instead of y and x. The resulting graph will be a straight line. Let us consider as an example y =» 2x*. The graph of this function was found in § 111 to be a parabola. The values given to x were 0, 1, 2, 3 . . . If we plot corresponding values of y and x* the table of values will be *• l 2 S 4 y 2 * 6 8 and so on. The values of x are not shown in the table. 241 2 4 2 TEACH YOURSELF ALGEBRA The resulting graph is a straight line passing through the origin as shown in Fig. 49. It is the same as y = 2x when y is plotted against x. If the equation is y = 2x 2 + 3 on plotting in the same way, that is y against x 7 , the resulting line will not pass through the origin but will have an intercept of 3 units on the y axis. It will be the same as y = 2x + 3, when y is plotted against x. m ■ ■■•»••••»••••■•• J »}■ ■■■■■■■■■■■■■■■■■••■ ■■■■•< aaaaaaaaaaa-.a-a-aa*! uiit ■■■■■■■••■••■•■•■■■■■■■a* m !■■■• ■■■■■■■■■■■■■■■■•■■■ 1 ml'.»« Wi !:::■■:::::■■:: ■ «*• ■■■■•■•••■ ■■!■■■'■■■ •■■■■a* •■>•■•• ■■••■ma ig !■■■■■■■■■■■« ••••*•••■•• )«■■■■■ ............ ••••• i! ii iaaataaaa*aa> ■••••••»•••• ■• ■■•■■■•' ■■■■•■■■■•■■■■■ «nii»>' «i ■■■■■■ ■*■■■•■■■ .■•.•••>•>>■•••••••*•••• ■if^Saa ■■■■■»■••■■■ ■■■>■■ liMIIKItlllllllllllllll ■■•■■•a ■•••••■ •■■•■■• ■aaiaaa ■■••■■•■■■1 ■•■■■■■■■■I ■■•■•••■•■< ■•■•(•■■■■i ■■■■■iiiiii ::::: ■■Kff::: MiMyi^ Hi ;^:::: ::::y:: ■.•.::ii:: Atuas •••■•• - ■■■■ HlMil ■ ■■l>l)ll>M !!•!■ :•;••;=■; ■*■• lillllilll^ k &i?M ■■::: :::;::::: li ;;;:;;;;;;:;:: ::.::::.:: is ::!»; ■■■■■■■■■ ========= ■■•• mm ........ :::::::: :::::::: :::::::::::: Fig. 49. The same procedure will be followed with any other power of x. Thus in y = 2x* -\- 5, y is plotted against x 3 . In general for the function y = ax" + b plot y against x*. If x" = z then the equation takes the form of y = az + b. This is the equation of a straight line. The graph, while not showing the relation between y and * as graphs usually do, will make it possible to find the values of a and b by methods previously given. THE DETERMINATION OF LAWS 2.43 Worked example. Two variables, x and y, are thought to be connected by a law of the form of y = ax' + b. The following values of x and y are known. Find the law connecting the variable. * 0-6 1 1-0 2 2-6 y -9-25 -7 -3-25 2 8-75 y must be plotted against ** ; we therefore calculate the following tables of corresponding values of x* and y. *> 0-25 1 2-25 4 6-28 y -9-26 -7 -3-25 2 8-75 ffi 2HS Sf'-'mftt* 111 i .1 ; 1 H _ ffi&Hjff 38 rr affi :'::• 5: I •• ||||||lH>f pfes 3± ItHlrtfl Fig. 60. Plotting y against x*. the resulting graph is as shown in Fig. 50. This is a straight line, and the values of a and b can be found by the method of § 168. By inspection of the graph, the intercept on the y axis {i.e., b) is —10, and a, the gradient of the line, is 3. :. the law is y = 3x a — 10. 244 TEACH YOURSELF ALGEBRA 181. y = ax". Use of logarithms. As was pointed out, the previous method can be used only when the power of x involved is known. If, however, it is not known, and the law is of the form y = ax", this can be reduced to the form of the equation of a straight line by taking logarithms. If y = ax" then logy = n log x + log a. Comparing this with the standard form of the equation of a straight line, viz., y = ax + b, it is seen to be linear and of the same form, log y taking the place of y, and log x taking the place of x. The constants to be determined are now n and log a. Therefore we plot the graph of logy = nlogx + log a. From this graph n and log a can be found in the same way as a and b in the standard form. When log a is known we find a from the tables and the law can be written down. It is possible to deal with only the simpler cases in this book. 182. Worked example. Two variables, x and y, are connected by a law of the form y = ax". The following table gives corresponding values of x and y. Find the law connecting these. X 18 20 22 24 25 | y 023 803 1100 1019 1724 Since the law connecting these is of the form y = ax" taking logs logy = n log x + log a. Tabulating values of log x and logy, we get the following: log* . 1-265 1-301 1-342 1-380 1-398 logy . 2-795 2-'.i:;:; 3005 3182 { 3-236 THE DETERMINATION OF LAWS 245 The logs correspond in order to the numbers in the column above, and are calculated approximately to 3 places of decimals. The graph is the straight line shown in Fig. 51. . ■»•■■■■■■■■■■■■■■■ <■■■■■■■■■ ■•■•■••■••■■■■•••■••■■■■■ !•■■■■■■■■■■■■■■■■■' ■■■■■■■■■•■■■■■■■•■■■...■... ■■■■■(! • ■■•••'iiiiiiittiiim ■■■■■■■■■■■■■•■■■>■■■ 27 ■''''''''l' T TTTt"tTffi | mm | 115 1-20 125 130 135 140 Log X Fig. 51. Selecting the points A and B on the straight line, we substitute their co-ordinates in turn in the equation logy = n log x + log a. Thus we get the equations 3-236 = l-398» + log a 2-795= l-255» -flog a. Subtracting, 0-441 = 0-143u. 0-441 " - 6T43 = 3-1 approx. 246 TEACH YOURSELF ALGEBRA Substituting for n in equation above 2-795 = (3-1 X 1-255) + log a. log a = 2-795 - 3-891 = 3-904. a = 0-08 approx. .*. the law connecting y and x is y = 0-OBx". Exercise 52. 1. The variables x and y are connected by a law of the form y = ax* + b. The following corresponding values of * and y are known. Find the law. X 0-6 1 1-5 2 2-5 y 4-5 9 16-5 27 40-6 2. The following table gives related values of x and y. Determine whether these values are connected by an equation of the form y = ax* + b and, if so, find the values of a and b. X 4 5 e 7 8 y 14-3 18 22-6 28 34-5 41-5 3. The following values of R and V are possibly connected by a law of the type R = aV* + b. Test if this is so and find the law. V 12 16 20 22 24 26 R 6-44 7-56 9 0-84 10-76 11-76 4. The horse power (H) which is necessary to drive a certain ship at a speed of V knots is supposed to be con- THE DETERMINATION OF LAWS *47 nected with V by an equation of the form H = aV 3 + b. The following corresponding values are known. Find the values of a and b. V . 10 12 14 15 17 H 1500 2300 3400 4000 5700 5. The following corresponding values of x and y were measured. There may be errors of observation. Test if there is a probable law y = a + bx* and, if this is the case, find the probable values of a and b. X i 1-5 2 2-30 2-50 2-70 2-80 y ■ 0-77 105 1-60 1-77 2-03 2-25 2-42 6. In measuring the resistance, R ohms, of a carbon- filament lamp at various voltages, V, the following results were obtained : V 00 70 80 1)0 100 120 R 70 67-2 65 63-3 62 60 The law is thought to be of the form R — p. -f- b. Test this and find a and b. 7. The values of x and y in the following table are con- nected by a law of the form y = ax*. Find a and n and so determine the law. * 2 3 4 6 y 2 6-75 16 31-25 8. The following table gives corresponding values of two X4» TEACH YOURSELF ALGEBRA variables x and y. The law which connects them is of the form y = ax*. Find this law. X 2 3 5 6 8 10 y 4-24 6-20 6-71 7-35 8-49 D-49 9. The following values of H and Q are connected by a law of the type Q = aH". Find a and ». H 1-2 1-6 20 2-2 2-5 3 Q • 6087 6-751 7-316 7-671 1 7-927 8-467 10. Two quantities x and y are connected by an equation of the form y = ax*. The following table gives correspond- ing values of the variables. Determine a and n. X 3 3-5 4 4-5 5 y 619 6-79 7-35 7-89 8-40 CHAPTER XX RATIONAL AND IRRATIONAL NUMBERS. SURDS 183. Rational and Irrational numbers. A number which is either an integer or can be expressed as the ratio of two integers is called a rational number. A number which cannot be expressed as an integer or as a ratio with a finite number of figures is called an I rratlonal number. Thus V2 cannot be expressed as a fraction or a ratio with a finite number of figures. It can, however, be ex- pressed as a decimal to any required degree of accuracy. Thus to 4 significant figures V2 = 1-414. „ 6 ,. „ V2 = 1-41421 and so on. But it can be proved that there is no limit to the number of figures in the decimal place, that is to say the number of figures is not finite. It is therefore an irrational number. Other roots such as V3, vTT, \/l9, tyi, ... are ex- amples of irrational numbers. A root of a rational number which is irrational as these are is called a surd. There are also other numbers, which do not involve roots which are irrational. Thus the ratio of the circumference of a circle to its diameter, which we denote by the symbol tc, cannot be expressed by a ratio with a finite number of figures. It is often expressed roughly by *,*, or 3-1416 to 5 significant figures, but modern Mathematics proves that there is no limit to the number of figures in the decimal part. One mathematician has worked out the value to 708 decimal places and he might have gone on for ever with the task. Such numbers as this are irrational, but not surds. They are also called Incommensurable. 184. Irrational numbers and the number scale. Since such numbers cannot be expressed with absolute accuracy it is not possible to assign an exact position for them in the complete number scale (§ 36.) We can, however, 249 2 5 TEACH YOURSELF ALGEBRA state to any required degree of accuracy the limits between which they lie. Thus V2 lies on the scale between 1-414 and 1-415 or more accurately between 1-4142 and 1-4143 or more accurately still between 1-41421 and 1-41422 and so on. 185. Geometrical representation of surds. It may be noted, however, that it is theoretically possible in many cases to obtain by geometrical constructions, straight lines which do represent surds accurately. For example, we know from Geometry that the square on the hypotenuse of a right-angled triangle is equal to the sum of the squares on the sides containing the right angle. Consequently if a right-angled triangle be constructed, the sides of which ar e of unit length, the length of the hypo- tenuse must be Vl* + I*— i.e., V2. Similarly the hypotenuse of a right-angled triangle of sides 1 , and V2 units will be V3 units. In this way it is possible to represent many surds by straight lines. The lengths thus obtained can be marked on the number scale, but in practice no high degree of accuracy can be obtained in the construction of them. 186. Operations with surds. It is the custom in Algebra to classify such a number as Va as a surd, though until a numerical value has been assigned to a we cannot say whether or no it is irrational. For purposes of operation, however, it is treated as a surd. In operating with surds one principle is fundamental. Surds must obey the laws of Algebra as formulated for rational numbers. Since surds can also be written as powers with fractional indices, e.g. V2 = 2*. we can operate with surds as with these powers, according to the laws of indices. Fo r exam ple, just as (a + b)' is not equal to a* -f b*. so Va + b is not equal to Va + VB- In this respect the root sign has the same effect as a bracket; the expression under it must be regarded as a whole. RATIONAL AND IRRATIONAL NUMBERS 251 (1) Multiplication. Va X Vb = a* x b* = (ab)* = Va x b. Thus: V3 x V7 = vTx~7 = V21 •v / 2(V5-V3) = (\/2x V5)-( V2x VS) = VIO- V6 4 V7 = VIBx VI = Vl6 x 7 = VH2 Zx*Vy = V9x* x Vy = VWy (Va" + Vb)* = (Va)» + 2(Va X Vb) + (Vb)* = a + b -t- 2Vab (Va - Vb)* = a + b- 2Va6 (Va +Vb)(Va- V>) = (Va)* - (Vb)* = a - b (VI + 7)(V3 - 2) = (VB x V3)-2V5 + 7V5-14. By using the above and the converse rules we obtain useful transformations in operations. Thus: Viooo= Vioox 10 = V100 x VTo = 10VT0 V72 = V86 X 2 = V36 x V 2 = 0V2 V9a s 6 a = V9 X a* x a X b* = ZabVa. The above transformations may also be employed to simplify expressions involving surds. Examples (1) VB + V20 = VB + V4~x"5 = Vs 4- 2 VB = 3 Vs. (2) V27 — V 75 + V 48 = V9 x 3 - V25 X 3 + VlG X 3 = 3V3 - 5V3 + 4V5 = 2V3- (2) Rationalisation. The evaluation of a number such as —t~ will be easier, and likely to be more accurate, if the fraction can be transformed so that we multiply by the surd and do not divide by it. This can be done by the following transformation: — = V2 = V2 V2 V2 X V2 2 ' 2 5 2 TEACH YOURSELF ALGEBRA Similarly 2V3 VI x V3 _ Vl6 Vl6 2 X 3 = 6 2V3 X V3 By this transformation the denominator is changed from an irrational number to a rational one. This is called rationalising the denominator. If the denominator is a binomial expression the method is slightly more difficult. The procedure is indicated in the following examples: Example I. Rationalise the denominator of —t= -m- Since (a — b) X (a + b) = a* — b' and (Va + Vb) x (Va — Vb) = a — b (see above), then, if the denominator is multiplied by V5 + V2, the surds will disappear from it. __!_ _ VB+ V2 VE-V2 (V5- VSUVS+VS) = V5 + V2 _ VS+ yi 5 — 2 3 Example 2. Simplify ^ -1 - To rationalise the denominator it must be multiplied by y/E-l. VS-i _ (VB-l)(V5-l) (V5-i)« '• VB + i (•v / 5 + l)(V5-i) (-s/5)*-(i)» 5 — 2V5 + l Thus 5- 1 6— J2V5 4 3-V5 Exercise 53. 1. Express the following as complete square roots, thus : 3V7 = VWxl = V53- (1) 5VB. (2) 12V2. (3) 10V5. (4) 4VT3. (5) 2aVb. (6) 3* a Vy. RATIONAL AND IRRATIONAL NUMBERS 253 2. Express the following with the smallest possible number under the root sign, in each case : (1) V800. (3) VW8. (5) y/ 375. (7) V2ia 3 b i . (9) VlBa'Pc*. (11) Va*b(a - W- 3. Simplify: (1) V3 x VTB. (3) V$2 x V24. (2) V320. (4) Vsooo . (6) v/ 7200. (8) \/ !8.tVz. (io) Viooopy. (12) V25%y»(* - 2y)«. (2) -v/14 x V7. (4) 2V5 x 5V2. (5) V2(2V2 - 1 ). (6) Vl(VU + V2). (7) Va + bx V a* - 6*. (8) V3(x - 2y) x V6(*» - 4y»). 4. Multiply the following: (1) (v/2-l)(V2-M) (2) (2\/3 + \/5)(V3-2V5). (3) (V2-l)». (4) (2VB+ V3)'. (5) (V7 - 5V5)". (6) (V5+ \/3)(V6- \/3). (7) (2-s/lO - V2)(2VlO + \/2). (8) (1 + 10V3)*. (9) (Va - 5)(V + 5). (10) (V27 + V&)(V2l - VB). 5. Simplify the lollowing by rationalising the denomina- tor: < 3) 2^7" ®VW (2) 1 vt 1 (4) sW (8) tf (9) 1 \/5- r (10) 3 V7 + vT *54 TEACH YOURSELF ALGEBRA (11) 2V3-V2" V3+ V2 (18) (15) Va- V2 3V7-1 (17) x/5 + V3~~V5^V3 (12) 2V7 - 3V3 (14) V^-A < 14 > V7 + V5 (16) ^Ti + VS+T (18) 6- V5 CHAPTER XXI SERIES Arithmetical and Geometrical Progressions 187. Meaning of a series. A series is a succession of numbers each of which is formed according to a definite law, which is the same throughout the series. The ordinary numbers 1, 2, 3, 4, . . . constitute a series, each term of which is greater by unity than the one which immediately precedes it. 5, 9, 13, 17, ... is a series, each term of which is greater by 4 than the one which immediately precedes it. 2, 4, 8, 16, . . . is a series in which each term is twice the one which immediately precedes it. 1, £, §, J, . . . is a series in which the terms are the reciprocals of 1, 2, 3, 4, . . . 1*. 2 a , 3« I s 2 3 3 3 are series, the construction of each of which is obvious. 188. The formation of a series. Series are of great importance in modern Mathematics, but in this book it is possible to deal with only a few simple cases. The two most important things to be known about a series are: (1) The law of Its formation. If this is known, it is possible to find any term in the series. (2) The sum of a given number of terms of the series. In this connection it is necessary to consider what is the nature of the sum when the number of terms is very great. 255 2 5 6 TEACH YOURSELF ALGEBRA If the series is one in which the terms increase numerically such as 2, 6, 8. 11. . . . or I, 3, 9. 27 . . . it is clear that the more terms which are taken, the greater will be the sum. But if the series is one in which the terms decrease as the number of terms increases, such as 1, J, 5 , S V, . . . , it is not always easy to discover what the sum will be when the number of terms is very great. This is a matter which will be considered later. 189. Arithmetic series. An Arithmetic series, or Arithmetic progression, is one in which each term is formed from that immediately preceding it by adding or sublraclittg a constant number. The number thus added or subtracted is called the com- mon difference of the series. Examples. (1) 7, 13, 19, 25, . . . (common difference 6). (2) 6, 4, 2, 0, — 2, . . . (common difference —2). In general if three numbers a, b, c, arc in Arithmetic pro- gression (denoted by A .P.) then b — a = c — b. 190. Any term In an Arithmetic series. Let a = the first term of a series „ d = the common difference (+ve or — vc) then the series can be written a, a + d, a + 2d, a + 3d, . . . It is evident that the multiple of d which is added to a to produce any term is one less than the number of the term in the sequence. Thus the fourth term is a + (4 — l)d = a + 3d. Hence if the number of any term be denoted by n then nth term = a + (n — l)d. Examples. (1) In the series 7, 10, 13, . . . the common difference is 3. .-. the tenth term is 7 + (10 — 1)3 = 34. • „ nth „ 7 + (n - 1)3. SERIES 257 (2) In the series 6, 2, — 2, — 6, ... d = — 4. .-. the nth term = 6 + (» — 1)(— 4) = 6 — (n — 1)4. .-. the eighth term = 6 -f 7(— 4) = 6 - 28 = - 22. 191. The sum of any number of terms of an Arithmetic series. Using the following symbols in addition to those used previously : Let n = number of terms whose sum is required „ s = the sum of n terms ., I — the last term then by the previous formula / = a + (n - l)d. Now s = a + (a + d) + (a + 2d) + . . . + (I - d) + I. Reversing the series s = /+(/- d) + (I - 2d) + ... + (a + d) + a. Adding the corresponding terms of the two sets, each pair gives (a + Z). 2s = (a +i) + (a + /) + («+*)+• ..+(*+/) + (« + /) = (a + [) x n, since there are n terms and .\ n pairs. s ~ 2 • Since / = a -{- (« — l)d, on substituting for / in the last result . n{a + a + (n - \)d) 5 ~ 2 s = !J{2o + (n- \)d}. This formula, like all other formulae, may be used not only to find s, but also any of the other numbers n, a, or d. To find a and d offers no difficulty, but if n is required it will be seen that a quadratic equation will result. Since there are two roots to every quadratic equation, two values of n will always be found. In some cases only one root is admissible; in others both roots provide solutions. 2 5 8 TEACH YOURSELF ALGEBRA For example, in a series involving negative terms such as 9+7+5+3+1-1—3 it is seen that the sum of 7 terms is the same as the sum of 3 terms. In other cases it will be evident that one of the roots is inadmissible. 192. Arithmetic mean. If three numbers are in Arithmetic progression, the middle one is called the Arithmetic mean of the other two. Let a, b, c, be three numbers in A. P. then by the definition of § 189 b — a = c — b 26 = a + c. b = a + c It will be seen that the Arithmetic mean of two numbers is the same as their average. It is usual also to speak of inserting Arithmetic means between two numbers, by which is meant that they, together with the two given, form a series of numbers in A.P. Example. Insert three Arithmetic means between 4 and 20. If these be a, b, c, then 4, a, b, c, 20 are in A. P., five terms in all. Using I = a + (n — l)d for the fifth term 20 20 = 4 + (5 — l)d, whence d = 4. .*. the five terms are 4, 8, 12, 16, 20. 193. Worked examples. Example I. The sum of an A. P. of 8 terms is 90 and the first term is 0. What is tlie common difference? Using s = s {2a + (» — \)d) and substituting given values 90 = S{(2 x 6) + (8 - \)d] 90 = 4(12 + Id) = 48 + 28a\ .-. 28a" = 42 and d = 1-5. <ERIES 259 Example 2. How many terms of the series 3, 6, 9, . . . must be taken so thai their sum is 135 ? Using s=|{2« + («-l)rf) and substituting, 135= g {6 + {n — 1)3). 270 = »(3 + 3») or 270 = 3n + 3m 1 . .-. n* + n — 90 = 0. Factorising (n — 9) (n + 10) = n = 9 or — 10. The root — 10 is inadmissible as having no meaning in this connection. .". the solution is n = 9. Exercise 54. 1. Write down the next three terms of the following series : (1) 5, 7-5, 10, . . . (2) 12, 8, 4, . . . (3) (a + 3b), (a + b), (a - b). (4) 2-7, 4, 5-3. (5) x — y,x,x+ y. 2. Find the fifth and eighth terms of the series whose first term is 6, and common difference 1-5. 3. Find the 2£th term of the series whose first term is 6 and common difference 2. 4. Find the «th term of the series whose first term is {x + 2) and common difference 3. 5. Find the twenty-fifth term of the series 0-6, 0-72, 0-84, . . . 6. The fourth term of an A.P. is 11 and the sixth term 17. Find the tenth term. 7. The fifth term of an A.P. is 11 and the ninth term is 7. Find the fourteenth term. 8. Which term of the series 23, 4-2, 61, ... is 36-5? 26o TEACH YOURSELF ALGEBRA 9. Find the sums ol the following series: a) 15, 16-5, 18, . . . to ten terms. b) 9, 7, 5, ... to eight terms. c) 0-8, 0-6, 0-4, ... to nine terms. d) 2f , 3J, 4$. ... to twenty-seven terms. 10. How many terms of the series 10, 12, 14, . . . must be taken so that the sum of the series is 252 ? 11. How many terms of the series 24, 20, 16, . . . must be taken so that the sum of the series is 80 ? 12. Find the thirtieth term and the sum of thirty terms of the series 4, 8, 12, . . . 13. A contractor agrees to sink a well 250 m. deep at a cost of £2-70 for the first metre, £2-85 for the second metre, and an extra 15p for each additional metre. Find the cost of the last metre and the total cost. 14. A parent places in the savings bank 25p on his son's first birthday, 50p on his second, 75p on his third, and so on, increasing the amount by 25p on each birthday. How much will be saved up when the boy reaches his sixteenth birthday, the latter inclusive? 194. Harmonic progression. A series of numbers is said to be in harmonic progression (H.P.) if their reciprocals form a series in arithmetic progression. Thus the series 1, 3, 5, 7, . . . are in A.P. ■• I. i I h • • • ,. H.P. This series is important in the theory of sound. It is not possible to obtain a formula for the sum of n terms of an H.P., but many problems relating to such a series can be solved by using the corresponding arithmetic series. Harmonic mean. — The harmonic mean of two numbers may be found as follows: Let a and b be the numbers. „ H be their harmonic mean, i.e., a, H, b are in l^P. A P SERIES 261 x . 111 1 hen -, H , 7 are in a ti b we+3 •■• a +b 2ab lab H=« 195. Geometric series, or geometric progression. A geometric series is one in which tlie ratio of any term to that which immediately precedes it is constant for the whole series. This ratio is called the common ratio of the series. It may be positive or negative. Thus each term of the series can be obtained by multiplying the term which precedes it by the common ratio. Examples. (1) I 2, 4, 8 (2) 1. i I * (3) 2, - 8, 18, - 54, (4) R, R*, R*. R\ . . common ratio 2). ». -3). R). If three numbers a, b, c are in geometric progression (G.P.) then *>-={. a b This is the test to apply in order to find if numbers are in General form of a geometric series. Let a = 1st term. „ r = common ratio. Then the series is a, ar, ar 2 , ar 3 , . . . 196. Connection between a geometric series and an arith- metic series. In the geometric series a, ar, ar s , ar 3 , . . . take logs of each term. We then get the series : log a, log a + log r, log a + 2 log r, log a + 3 log r, . . . I (ALG.) 26z TEACH YOURSELF ALGEBRA This is an arithmetic series in which the first term is log a, and the common difference is log r. :. the logarithms of the terms of a G.P. form a scries in A .P. 197. General term of a geometric series. Examining the series a, ar, or*, ar 3 , ... it will be seen that each term of the series is the product of a and a power of r the index of which is one less than the number of the term. .". if n = any term then the nth term = ar"-'. If r is negative, n — 1 being alternately odd and even, the terms will be alternately negative and positive, assum- ing a to be positive. When n — 1 is even, n is odd, and the wth-term is -f-ve. „ « — 1 is odd, « is even, „ „ is — ve. Worked examples. Example I. Find the seventh term of the series 3, 6, 12, . . . In this series r = 2, so using the formula «th term = ar" " l the seventh term = ar 8 = 3 X 2' = 3 X 64 = 192. Example 2. Find the eighth term of the series 2, — 6, 18, — 54, + . . . For this series r = — 3. Using ar" ~ 1 , eighth term = 2 x (- 3) 8 " » = 2 x (-3)' = - 4374. Example 3. Find the fifth term of the series in which the first term is 100 and the common ratio 0-63. Using ar*~ l ,iix = the fifth term x = 100 x (0-63)'. log x = log 100 -f 4 log 0-63 = M972 = log 15-75. the fifth term is 15-75. No. ! log. 0-63 100 15-75 1-7093 4 1-1972 2 11972 SERIES 263 Example 4. The third term of a G.P. is 4-5 and the ninth is 16-2. Find the common ratio. Using ar" - *, third term = or* = 4-5 ninth term = ar s = 16-2. Dividing Taking logs ar 9 -4- ar % = 16-2 -=- 4-5. r« = 16-2 -4- 4-5. 6 log r = log 16-2 - log 4-5 = 0-5563. No. log. log r = 0-5563 -^ 6 •m 00927 16-2 1-2095 - log 1-238. 4-6 0-6532 r =. 1-238. 0-5563 198. Geometric mean. If three members are in G.P., the middle term is called the geometric mean of the other two. Let a, b, c be three numbers in G.P. Then by the definition of § 195 b_c a~b' b* = ac,_ and b = Vac. Exercise 55. 1. Write down the next three terms of each of the follow- ing series : a) 4, 10, 25. b) 16, 4, 1. c) 16, — 24, 36. d) 0-3, 003, 0003. e) 3, 0-45, 0-0675. 2. Find the seventh term of the series 5, 10, 20, . . . 3. Find the seventh term of the series 6, — 4, 2§, . . . 4. Find the fifth term of the series 1-1, 1-21, 1-331, . . . 5. Find the sixth term of the series — 0-5, 0-15, - 0045, . . . 199. The sum of n terms of a geometric series. In addition to the symbols employed above, Let S„ represent the sum of n terms of a G.P. then S n = a + ar + ar* + . . . + ar"-* + ar"- 1 Multiply both sides by r, then rS n = ar + ar* + ar* . . . + ar" ~ l + ar" Subtracting (1) from (2), rS H — S n = ar" — a or S n (r - 1) = a(r" - 1). (1) (2) °» — r _ I r — (A) 264 TEACH YOURSELF ALGEBRA 6. Write down the 2«th and the (2n + l)th terms of the series: (1) a, ar, ar*, . . . (2) a, — ar, ar*, . . . 7. The first term of a G.P. is 1-05 and the sixth term is 1-3401. Find the common ratio. 8. The fifth term of a G.P. is 1-2166 and the seventh term is 1-3159. Find the common ratio. 9. Find the geometric mean in each of the following cases: (1) 3 and 5. (2) 4-2 and 3-6. 10. Insert two geometric means between 5 and 13-72. 11. A man was appointed to a post at a salary of £1000 a year with an increase each year of 10 per cent, of his salary for the previous year. How much does he receive during his fifth year ? 12. The expenses of a company are £200,000 a year. It is decided that each year they shall be reduced by 5 per cent, of those for the preceding year. What will be the expenses during the fourth year, the first reduction taking place during the first year ? 13. In a geometric series the first term is unity and the fifth term is 1-170 approx. Find the common ratio. 14. Insert three terms in geometric progression between 5 and 80. SERIES 2 6 5 If (2) be subtracted from (1) above, the formula becomes: — r (B) If r > 1 and positive, form (A) should be used. If r < 1 or negative, form (B) should be used. Note. — It will often be necessary to employ logarithms to evaluate r". They cannot of course be employed to evaluate the whole formula. 200. Worked examples. Example I. Find the sum of seven terms of the series 2, 3, 4-5, . . . r = f = 1-5. Using S=^i>, and substituting S = 2( 1 1 '?' ~^ 1-5 — 1 logs may be used to find 1-5'. Let x = 1-5 7 . then log x = 1 log 1-5 = 7 x 01761 = 1-2327 = log 17-09. 1-5' = 17-09 (approx.). ^ = 4x16-09 = 64-36 = 64-4 approx. 0-5 Example 2. Find the sum of seven terms of the series 4, - 8, 16, . . . r=— 2. Using 1 —r and substituting S = ^ ~ (~ 2 H = 4 C + 128 > 1 ( 2) 3 2) i X 129 = 172. 266 TEACH YOURSELF ALGEBRA SERIES 267 Exercise 56. 1. Find the sums of the following series: (a) 1-5, 3, 6, ... to six terms. (6) 30, — 15, 7J, . . . to eight terms. ( c ) i. — i. J. • • • to six terms. 2. Find the sum of the first six terms of the series 5, 2-5. 1-25, . . . 3. Find the sum of the first six terms of the series 1 - 1 + * 4. Find the sum of the first six terms of the series 1 + 1-4 + 1-96 + . . . 5. Find the sum of the first twelve terms of the series 4 + 5 + 6-25 + . . . 6. If the first and third terms of a G.P. are 3 and 12, find the sum of eight terms. 7. If the third and fourth terms of a G.P. are £ and J, respectively, find the eighth term and the sum of eight terms. 8. Find the sum of 20 + 18 + 16-2 + , . . to six terms. Infinite Geometric Series 201. Increasing series. When the common ratio of a geometric series is numeri- cally greater than unity, as in the series 1, 2, 4, 8, . . . 2-5, 7-5, 22-5, . . . the terms increase in magnitude. The sum of n terms increases as n increases. If the number of terms increases without limit — that is, n is greater then any number we may select, however great — then the sum of these terms will also increase without limit, i.e. it will become infinitely great, or, to use the mathematical term, approach " in- finity ", which is denoted by the symbol co. We may say, then, that as «, the number of terms, approaches infinity, S„, the sum of these terms, also approaches infinity, ing notation. If n- This may be expressed by the follow- »■ 00 , then S„ — > eo . 202. Decreasing series. If, however, the common ratio is numerically less than unity, as in the following series, 3' i' tV> • • • 0-2, 002, 0002, . . . then as the number of terms increases, the terms themselves decrease. Using the terms employed above, we may say that, as n increases without limit, the terms themselves decrease without limit, and ultimately become indefinitely small. We cannot say, however, that the sum of these terms increases without limit, as n increases without limit. That is a matter for further investigation. 203. Recurring decimals. There is an example, arising from Arithmetic, which will assist in coming to conclusions on this important question, viz. that of a recurring decimal. We know that in which 1 recurs without limit. The decimal is in effect the geometric series 1.11 .+ 108+ • • • 10 + 10» in which there is no limit to the number of terms. It is an example of what is called an infinite series. But we know that the sum of all these terms, no matter how many are taken, is ultimately equal to the finite fraction £. If we find the sum of finite numbers of terms, we get: 3i - 10 - S -I + -J-- -'! — in ' 1A2 10 _1_ 10 2 1 11 1 111 S» = TS + ip + Jos = 1Q3 and s0 on - 268 TEACH YOURSELF ALGEBRA The difference between J and the sums of these is: i c 1 y -f i — tb 1 C 1 5 — °S — 571 or and in general, finding the sum of n terms by using the formula SERIES 269 1— r we get S.= IoA 10V -fo 9V 10-/ ~ 9 1 9 x 10»* Examining these results, it is seen that the difference between ^ and the various sums, S j, S 2 ,S S , . . . S„ decreases as n increases. In general, the difference between J and the sum of n terms is s T7r . 9 x 10" As n increases without limit, this difference decreases without limit — i.e., it tends to become zero — and the sum approaches to equality with J. It can never be greater Using the previous notation, we can express the result thus as n — > oo , S„ — > $. There is thus a limit to which S„ approaches and which it cannot exceed. 204. A geometrical Illustration. The approach of the sum of a geometric series to a limit may be illustrated by a graphical representation of the series 2+4 + 8 + 16+ • • ' or 1 . I I + 92 + 03 + 04+ • • Let the rectangle A BCD (Fig. 52) represent a unit of area. HD 8 F H L Fig. 52. Let E be the midpoint of AB and draw EF perpendicular to DC. Then rectangle AEFD represents | of a unit. Bisecting the rectangle EBCF by GH, then rectangle EGHF represents ■? or 55 of a unit. Continuing the process of bisecting the rectangle left over after each bisection, we get a series of rectangles whose areas represent the terms of the above series. These rectangles diminish in area as we represent more and more terms of the series in this way. The rectangle AKLD represents the sum of the four terms of the series 2 ^ 2 2 ^ 2 8 ^ 2 4 ' As more divisions are made and the sum of more terms represented by a rectangle, this rectangle approaches nearer to the area of the whole rectangle^ — i.e., 1 — but can never exceed it. Consequently 1 is a limit which the sum of the series approaches as the number of terms is increased without limit, but which it can never exceed, no matter how many terms are taken. If the series | + ^ + 55 + • • • be summed by using the formula S„ = \ _ ' , weget 5b = «^B_tf^!B3. s„ = 1 - (J)". Examining this result, we see that (J-)" decreases as n TEACH YOURSELF ALGEBRA 270 increases. If n be increased without limit then (J)" de- creases without limit — i.e., it approaches zero. /. we can say that S„ — > 1 as n — >■ 00 . 205. The sum to infinity. The above suggests the general treatment of this question. _ a(l — r") 1 — r a — ar" Using i.e., we have S„ = S„ = 1-r a 5 « = rf_- r -«-— ,• r n Considering the term a . y^—, if r is a proper fraction {i.e., it lies between + 1 and — 1) then r n diminishes as n increases or, with the previous notation, as «• co, r" and a . 1-r a ->-0. Thus the right hand side approaches a _ as a limit. This is the " limiting sum " of the series and it is called the " sum to infinity ". If it be represented by S B then s_ = — r 206. Worked examples. Example I. Sum to infinity the series 2 + i + J+ . . . Here a = 2, r = \. S„ = -t 1 -2-J-i Example 2. Find the sum to infinity 0/ the series 5-1+i- . .. SERIES Here a = 5, r = — J. 271 -5r^ Exercise 57. 1. To what limits will the following series tend as the number of terms increases indefinitely? f &+ - - - 2. To what limits will the following series tend as the number of terms increases indefinitely ? (o) 01 + 0001 + 000001 -f • • • (b) 0-06 + 0-0006 + 0000006 + . . . (c) 0-16 + 00016 + 0000016 + . . . What is the connection between these series ? 3. To what limit does an infinite number of terms in the following series tend ? l-i+i-i+-.. Show the connection with the series in question 1. 4. Show that the sum of « terms of the series is «1 - (*)"}• Hence show what limit this series approaches. 5. Find the sum of « terms of the series when a u Hence find the limit approached by the series as the number" of terms becomes infinitely great. : 7 2 TEACH YOURSELF ALGEBRA 6. Find the limiting sum, or the sum to infinity of the following series : I + i + A- + 6)5-1+$- (c) 9 - 6 + 4 - . . . 7. Find the sum to infinity of the series: • o' T £? j UI ? t0 infinit y of a se "es is 15, and the first term is i. tind the common ratio. 9 A rubber ball is dropped from a height of 16 ft. At each rebound it rises to a height which is f of the height from which it has just fallen. What is the total distance through which the ball will have moved before it finallv comes to rest ? J 10. The yearly output of a silver mine is found to be decreasing by 25 per cent, of its previous year's output If in a certain year its output was £25,000 what could be reckoned as its total future output ? 207. Simple and compound Interest. The accumulation of money when put to interest furnishes examples of arithmetic and geometnc series. When money is put out at simple interest, the interest is payable for each year, but is not added to the principal. / F , or „ ex „ am P le the interest payable on £100 at 5 per cent. for 1, 2. 3 . . . years will be £5, £10, £15 these sums forming a series m A.P. and the interest varies directly as the time. ' • B jj 'i mo " e y is lent a t compound interest, the interest is added each year to the principal, and for the following year the interest is calculated on their sum. Suppose £1 to be invested at 5 per cent, compound in- terest. Then the interest for the first year is £J^ or £0-05. .-. the amount at the end of the year is £1-05. and " of£Pis£Px 1-05. SERIES a 73 Consequently the ratio of the amount at the end of a year to that at the beginning is always I -05. This corresponds to the ratio of a geometric series. .•. amount at the end of the second year is P X 1-05 X 1-05 = P X 105*. .*. amount at the end of the third year is P X 1-05* X 105 = P x 10o 3 . ,\ amount at the end of the fourth year is P X 105 3 X 105 = P X 105*. .". amount at the end of the nth year = P X l-05\ These amounts at the end of successive years, viz. I P X 1-05, P X 1-05*. P X 1-05 8 . . . constitute a geometric series. Let M = the amount at the end of » years then M - PR". In this formula, as we have seen in other cases, any one of the four quantities may be the subject of the formula. Thus r A"' This enables us to find the sum of money which will produce £M in n years. Again "=# from which the rate of interest can be discovered. Again n log R = log M — log P. log M — log P logR whence the time taken for P to amount to M is found. 274 TEACH YOURSELF ALGEBRA 208. Accumulated value of periodical payments. Suppose that £P is invested each year for 10 years at 5 per cent. C.I., each investment being made at the begin- ning of a year. Using the above formula: The first £P at the end of 10 years amounts to P x 1-05 10 . The second £P at the end of 9 years amounts to P X 105». The third £P at the end of 8 years amounts to P X 1-05 8 . And finally, the last £P invested bears interest for 1 year and amounts to P X 1-05. Then the accumulated value of the investments amounts to P X 1051° + P x 1-05" + Px 105 8 . . . +Px 1-05, or, reversing the series, P x 105 + P x 105 2 + Px 105* . . . +Px MB*. This is a geometric series and using the form in_ r-l the accumulated value of the investments is I05(I05"»- I) 105 - I xP. 209. Annuities. An annuity is a series of equal annual payments extending over a specified number of years, or for the life of the annuitant. Aground rent is a similar financial transaction, the holder of the freehold receiving an annual payment, called ground rent, for the number of years specified in the lease. Ground rents and annuities are constantly being bought and sold, and the method of calculating the amount to be paid by the purchaser can be determined by means of the above SERIES 275 results. This amount will depend upon the rate of interest which the purchaser expects to receive on his investment. Let the rate of interest expected be 4 per cent. The price is obtained by finding the present value of each of the payments, as follows: From the formula M = PR" we get as shown r R-- P is the amount which produces M in n years at the given rate per cent.; it is called the present value of M due in n years. If £A be the annual payment and P be its present value, then for the first payment due in 1 year P-A-A. C ~R~ 104 when the rate per cent, is 4. For the second payment P = For the third payment P = and so on. .'. the total present value A A 104*' A 1-04 3 ' •04 + 1-0 ="{rJ 04 + 1-04 A 4- 1-04 3 + 1 i + 1U4 3 + ...} = A 104 1 1 (.1-04/ j 1 - 1-04 0-04 [ 1-04-J This can then be evaluated for any value of A . The terms of the above series decrease and if the ground rent is a perpetual one, or the lease is a very long one, the 27« TEACH YOURSELF ALGEBRA present value becomes the sum to infinity of the above series — i.e., 1 104 Present value = A 1 - 1 104 ■A 1 104 1-04-1 104 MffH-^T-'* 25 - The ground rent is then said to be worth 25 years pur- chase. Tt is always found by dividing 100 by the rate per cent. Exercise 58. 1. If £100 be invested at the beginning of each year for 10 years at 3 per cent. C.I., find the accumulated value a year after the last amount is invested. 2. An annuity of £600 a year is allowed to accumulate at 3 per cent. C.I. for 8 years. What was the total amount at the end ? 3. A man saves £25 every half-year and invests it at C.I. at 4$ per cent. What will be the amount of his savings in 8 years if the last amount saved bears interest for 6 months ? 4. Find the present value of an annuity of £300 for 10 years, reckoning C.I. at 4 per cent., the first payment being due one year after purchase. 5. What should be the purchase price of an annuity of £500 for 8 years, reckoning C.I. at 3J per cent. ? 6. A man wished to endow in perpetuity an institution with a yearly sum of £200. If C.I. be reckoned at 4 per cent., what amount will be needed for it? 7. A man retires at 65, when the expectation of life is 10-34 years, with a pension of £200. What single payment would be the equivalent of this, reckoning C.I. at 4 per cent. ? 8. A pension of £6000 per annum was awarded to Nelson and his heirs for ever. If this be commuted into a single payment, what should that be, reckoning C.I. at 21 per cent.? APPENDIX The following brief statement ol Permutations, combina- tions and the Binomial Theorem, and a note on the roots of a quadratic, equation are given for the benefit of students who may need to use them in the Differential Calculus or other branches of more advanced Mathematics. Permutations and Combinations I. Permutations. Consider the following example: A party of 6 people arrived at a theatre and obtained 4 •eats together and 2 separate. In how many different ways could the 4 seats in a row be filled if there are no restrictions as to where any of the 6 may sit ? Consider the first seat. Since any one of the 6 people may sit in it, it can be filled in 6 different ways. With each of these 6 ways, the second seat can be filled in 6 different ways, since 5 people are left to choose from. ,\ there are (6 x 5) different ways of filling the first two teats. With each of the 6 x 5 or 30 ways of filling the first two seats, there are 4 ways of filling the third seat, since 4 people are left to choose from. .". there are (6x6x4) different ways of filling the first three seats. Similarly the fourth seat can be filled in 3 ways and .*. there are (6 X 5 X 4 x 3) different ways of filling the 4 seats — i.e., :$00 ways. Arrangements of a number of different objects in a row are called Permutations and the above problem was that of the permutation of 6 things 4 at a time. This is expressed by a special notation— viz., a P 4 or e P 4 . It will be seen that if the people mentioned above had 6 seats together, the number of permutations or arrangements in these seats would be 6x5x4x3x2x1. This product of all the integral numbers from 1 to 6 inclusive is called Factorial 6 and is expressed by the j 6 or 6 I K ikuu 2?7 278 TEACH YOURSELF ALGEBRA and In general, the product of the integral numbers from 1 to n inclusive is denoted by |« or n I Thus |»=n(» — 1)(» — 2j ... 3, 2, 1 and is called Factorial n. 2. Permutations of n things r at a time, or "P r This is the general treatment of the above special case and the method adopted to find the formula is the same. There are r places to be filled and n different things to choose from. The 1st place can be filled in n ways .. 2nd „ then be filled in (n — 1) ways since with each of the ft ways of filling the 1st place, each of the (n — 1) ways of filling the second can be associated. .*. there are n(n — 1) ways of filling the first two places. Similarly there are n(n — l)(n — 2) ways of filling the three places, u(n — l)(n — 2)(« — 3) ways of filling the four places. .*. by inspection there are n(n - 1)(» — 2)(m - 3) ...{«— (r - I)} ways of filling the r places. .-. "P, = n(n - l)(n - 2)(n - 3) . . . (n - r + I). If the w things are all arranged among themselves, then the last factor becomes (« — « + 1) or 1. "Pn = In. 3. Combinations. The problem solved above— viz., the number of different ways of filling up 4 seats by 6 people— might have been approached in another way. (1) We could find the number of different sets or groubs of 4 that could be formed from 6 people. (2) Each group could then be arranged in the seats in [4 ways. The product of these two numbers must give the total number of ways of filling the 4 seats— i.e., the permutations APPENDIX 279 of 6 things 4 at a time. The difficulty at present is that of finding the number of groups. Let x = the number of groups then, by the above reasoning, x = 14 - «P 4 . X - % -r [4. Thus we can find the number of groups, when we know the number of permutations. Such groups are called Combinations and a notation similar to that used for Permutation is employed to express them. Thus the number of groups or combinations of 6 different things 4 at a time is denoted by *C 4 , and in general, the number of combinations of n things r at a time is denoted by "C, or «C r . 4. The Combinations of n things r at a time. With the same reasoning as that employed above we can deduce that »C r « *P t 4- If «(n — l)( w — 2) . . . (n — r + 1). or t »c r = '-y Thus "C 2 = !*L^i> BC3= n(n-l)(n-2) lor _ 10X 9x 8 X 7 _ olft °« _ 1 X 2x3x4 ~- 1U 8r _ 8x7x6 _ „ Cs ~ 1 X 2 x 3 = 50 - The Binomial Theorem 5. Products of binomial factors. It was shown in § 77 that (* + a)[x + b) = x* + x(a + b) + ab. j8o TEACH YOURSELF ALGEBRA Employing the methods used in the chapter we can show that (x + a)(x + b)(x + c) = x 3 + x*(a + b + c) + x{ab + be + ca) + abc. It should be noted that: (1) The expression is arranged in descending powers of*. (2) The coefficients of these powers, after the first, are the sets formed in every way of the letters a, b, c, (a) one at a time, (b) 2 at a time, (c) 3 at a time. From the way in which this product is formed we can deduce the product of (x + a)(x + b){x + c)(x + d). Arranging the powers of x in descending order, the coefficients of these powers will be x* — unity. x* — sum of the letters one at a time, i.e., (a + b + c + d). x* — sum of the letters two at a time, i.e., (ab + bc + ad + bd + cd + ac). x — sum of the letters three at a time, i.e., (abc + bed + acd + abd). Term independent of * is abed. .'. the full product is x* + x*{a + b + c + d) + x*(ab + be + cd + ad + bd + ac) + x(abc ■+ bed -f- acd + abd) + abed. We have seen from § 4, that the number of ways of (1) Grouping 4 letters 1 at a time is 'C, = -r 4. (2) M 4 l» 2 II n l C t 4 " 1 3 (3) *• 4 II 3 II it *c 3 4 1 8 2 2 .3 (4) ll 4 II 4 II it f c t = 1. s, = 4. In the above factors let b = c = d then the left side is (x + a)*. APPENDIX *8i In the expansion of it : The coefficient of x* is (a + a -f a + a) = 4a. .. .. x* „ 6a«. .. * .. 4a 3 . The last term is a*. .-. (x + a) 1 = x 1 + 4x*a + 6x»a» + 4xa» + a* or (x + a)* = x* + K x x*a + «C,xV + «C 8 xa» + a*. By a similar process we can obtain the expansion (x + a) 5 - x« + Kjpa + 6 Cj?a* + *C# % # + i C i xa* + a». From a consideration of these results we can deduce the general case — viz., (* + a)" = x- + "C^xr-H + -tV" *a s + -CjX"-»a 8 + . . . + a* or (x + a)" = x" + nx"- >o + 5fej^H*»-%" + n(n-l)(n-2) x „_3 , + +<| . L? This is called the Binomial Theorem. The above reasoning is independent of the values of x and a. It will therefore hold if a be replaced by —a. Then (x - a)" = x" + «*»- l (- «) + HTT^ *" " 8(_ a)t + Since odd powers of (— a) are negative and even powers are positive, the terms will be alternately +ve and — ve. .-. (x - a)» - x* - nx*~ l a + "^ ^ ■ ** -*n* - »(»- 1 H» 3 - 2 > x"-»a»+ ...+(—)-■ In the above results let x = 1. 282 Then TEACH YOURSELF ALGEBRA (1 + «)— 1 + na + flfell a' + »<» -M- 2 > a » + " . . . + «■ and (1 — )■ - 1 - na + g^D a» - »<» -'»»,- 2 > a» + '..'.+ (- «)". Every binomial expression can be reduced to one of these forms as follows: Similarly, A complete proof of the Binomial Theorem requires a more advanced knowledge of Algebra than is provided in this book. The demonstration given above assumes that n is a positive integer. In using the theorem later the important question will arise: Is the Binomial Theorem true for fractional and negative indices ? It can be proved that the form holds for all values of n. For example of n = J. When n is a positive integer, a term in the series will ultimately be reached when one of the factors n(«- l)(w — 2) . . . will become (« — n) and will vanish, as will all succeeding products. The number of terms therefore is finite and will clearly be n + I. APPENDIX 283 Consequently when a value is assigned to x in (1 + x) n the sum of the series will be a finite number and the series is said to be convergent. But if n is fractional or negative, none of the factors n(n — l)(n — 2) . . . will vanish, and the number of terms is Infinite. If the sum increases without limit as n increases the series is said to be divergent. But it can be shown that if x, in (1 + x) n , is numerically less than unity, the sum of the series will approach a limit, as is the case with certain geometric series (see § 205). Subject to this condition, the series is always convergent. The Roots of a Quadratic Equation Writing the general quadratic equation ax* + bx -f- c = b c in the form x* + - x + - = . . . . (A) the solution is == - dL± VF=lac (§127) 2a Let a and (5 be the two roots „ , „ -b+ Vb*-4ac , -b - Vb* - 4<ic then a +P = ^ + g _26 = _b = 2a a' = coeff. of x In equation (A). . . „ - b + Vb* — 4ac —b- Vb* - 4<w Again a3 = — — ^r~ - X 2a -- (- b)* - (Vb* - 4ac)« 4a » (§83) = term Independent of x In (A). Summarising, a + B = aB (cf. § 77) 284 TEACH YOURSELF ALGEBRA Nature of the roots of a quadratic equation. If (b 2 — 4ac) is negative, the square root has no arith- metical meaning. It is customary to speak of such a root as Imaginary, while the square root of a positive number is called real. Hence, for the roots of a quadratic. If b 2 > 4ac, the roots are real and different. 21 If b 2 = 4ac „ „ ,, equal. (3) If b 2 < iac „ „ „ Imaginary. Such a number as V— p can be written Vp X (- 1) = Vp X V^~l. The number V— 1 is usually denoted by t. Hence, as an example, V— i> can be written ± V% x — 1 = ±3». Example. — The roots of the equation x 2 + 2x + 5 = are given by - 2 ± Vi^To x- ^—j- _ _ 2 ± V- 16 2 _ - 2 ± 4» _ 2 = - I ± 2/. Note.— It will be found that the graph of *• -f 2x + 5 does not cut the x axis, ».«., it does not equal zero lor any real value of x. p. 20. 1. (1) 100*. 100. n 4. (1) 1760a. 5. 28 — n. 100* 36 ' 9. 2n + 5. xa +yb ■ 100 * 13. a-b. 15. a + 6. 17. m/> + »?• 8. (1) 19. xv\ p. 27. t s 3. 3a 6. (1) •2. ANSWERS Exercise I. « iff- (2) 5280- 3. lOOan + mb. (3) 20' 7. 2. x 6. * + 60, or 50 — x (2) IZfPJLW. 10. (1) * + 2, * - 2. (2) * + 1, * - 1. 12. * + & Gy 14. *y. 16. * + 2 y - ,V 18. 100a + 6. 20 I. »i — 2, n — 1, «, n + 1. n + 2. Exercise 2. 6. 7. 8. 9. (1 (1 (1 (1 (3) 10. 3|. 12. (1) 13. 3ji 15. a, 16. 2m (2) lOdoz.; 120. 5x73 = 366. + 46; 47. 266. 2a + 46. 10a + 26. 13. 8a6 = 64. 3xy ■= 45. (6) 10a; 120. (6) 56 = 365. 4. 19a; 475. 19*. (3) 12a. 2) 12/. + 3?. (3) 2) 26 + c - d. (3) (2) 7. (3) 13. 2) 3a* + 6* (4) 4*. 6a + 36 + 4. 8* + 2*. (4) 11. = 42. 36. 4) 4a6 + 46* — ay 11. 22. 66. (2) 29. (3) 30. + 12. 14. 15a. a + d, a + 2d. a + 3d, a + 4d; 5a + 10a\ + 5, 2n + 8, 2» + 11, 2« + 14, 2» + 17; lOn + 55. 285 286 TEACH YOURSELF ALGEBRA p- 31. 1. 1 2a. 4. SIMM 7. tiOabc 10. a*. 13. 2a\ 16. 66*. 19. 6a«. 22. Sa». 25. 54. 28. 172. 31. 216. p. 33. 1. a«. 5. 2a«. 0. ab. 13. 2. 1. t£ 3. 6. 7. 0. 11. Exercise 3. 2. 10*>>. fi ab 5 " 12- 8 24- 11. x*. 14. 6*' 17. x*y'. 20. 27a«6 s . 23. 16a'«. 26. 21. 29. 33. 32. 576. Exercise 4. 2. o». 3. 3*. 6. 5y». 10. *y 14. J. 37. 12a 35' 43-jfJJ* xy 11* 40/ 20a» + 96Je 24a«6c* ' 6a 6* + 5 ~~26" • a* + a + 1 7. 3*». 11. oa'-b. Exercise 5. 6. 5& 3. 2xy. 6 4a6. 9. *». 12. «*•. 15. 2a'6«. 18. 7*«a'. 21. *•. 24. «4a«. 27. 8. 30. I. «■?■ 8. 2e«. 12. 3*y. 2 - *' 15- . 19a *■ 6F- 31* 18/ 3 ** + 4 3* - 10 WZL« 8. 12. 6y r ** — *+! ,, c+2a+ 36 "■ abc ■ 14. 56c - 4ac + 2a6 a J 6«c« 15. *«y. 18. 1. a* 19. *y. 17 V '' I7 - 2aV 20'. 6e 21. 35- 24. % 27. 1. 2x* &■ 30. » %■ 36. 4« 2 6« ANSWERS 22 3y T 23 9 •V 2;, 2*« T 26. ly. 2S X* 29. 1 31 X* 32 2a» 36* M 3y« 35. 20 4 ' ^y Exercise 6. 287 p. 43. 1. 15* + 18*. 2. 6a« + 8<j6. 3. 18a" + 42a s 6 - 36a 2 c. 4. 8* + y. 5. 5** + 4*«. 6. *. 7. 8* + 5,y - 4». 8. * - 2y - *. 9. 2* - y + 2*. 10. a + 76. 11. a -6. 12. a + 6. 14. a + 56 - 5c. 15. 3j-. 17. ** + x*y — x*y + xy*. 19. 9/>» + 3/>? + 16?*. 21. 2x* + Sx*. 13. 4* + 2v — 2*. 16. a* + 6«. 18. *• + 9* + 23. 20. 5*y - Zxy + 6x\ p. 45. 1. 6a - 9. 4. 3a f + 2a6 — 2ae. 6. ** + Sxy + 12y*. 8. 13* + 5. 11. 2a- 106. 14. 17. 20. 2a - (6 - e). x t _ y( x _ y). (1) 111. Exercise 7. 2. 3a + 306. 3. §. 5. 2*« + p. 7. 76c - 26' + 2c». 34 _ 6a + 46. 10. * -I 3y. 3** ., 27c x - (v + *). 16. 2(a + 26 - 3e). 2a - 46 + 7c. 19. * + 2y - 2x. 6. 9. 12. 15. 18. (2) p. 53. Exercise 8. + 4-2 + 4-6 = 0. + 8 + 2-14 + 4-6 = (1) 1: is 11°. (2) + 5. » (a) +2. U 1+4 (2) (-4). V>) 0. (8) = - 0. 15°. -6. + 4. (+8). + 3). (-8.) 30°. + 8. + 5. « i88 7. TEACH YOURSELF ALGEBRA (8) - 8. 1) la. 4) - x - 8y. 7) 7* - 3y. 8. (1) 2x - 2y. I) 3a - (- 6a). 3) - 3a - (- 10a). 1) (- a). (2) (+ a). 0. 10. (2) - 7*. (3) 10a6. (6) a - 76. (si _ * + 2y. (2) 2* - y - 7*. (2) 5* - (6*). p. 57. Exercise 9. (2) (- 30). (5) (+ 4). 8) (+ 4). (2) (4 a*). - 10**. - 2ab*. + 20a6*. 13) (- a). (3) (" 36). W (- 4). (3) (- 1). (3) (- 20xy). -6». - Ixt + 2yx. a*. — a s , a 4 . — a B . 4*', — 8*», 4- 16**. - 32*» 6« 6» , b* 6» ., — a* + 2ab + ao. [2) 6*» + 8*. y 27' + 81' 243' ; 9. ± 3* s . - *, - 2a». ' +16*. (2) 5*. - 26. (6) - 2*. 8**. (8) - 3*. 6a'c. 10. (1) -b. p. 64. 1. (a) 2}. 2. a) 16. 3. (a) - l(i 4. a) 80. 6. la) 12-6. 6. a) 3J. 7. a) 2-6. 8. a) 8. 9. a (2) - *. Exercise 10. (3) iy. (6) 2*. (9) - 4a6. (3) - |- 10. 8. b) 20. 6) 60. 6) - 10. lb) - 72. - 008. 5. !*>) 12. !.. (6) - 45. ANSWERS *8j 11. (a) 148. 12. (a) V- (6) 181. H 1 13. 22. 14. 6. 15. 8. 18. 29. 17. -f 18. - 3. 19. 18*. (6) 27£ 20. (a) V- 21. (a) - 80. (6) - 10. 22. - 35. *&. it* 23. 8. 24. - 7. 26. J. 26. 3. 27. * 28. - V- 29 - 2 ' 7 - 30. 2-9. 31. 32. f*. 33. 36. 34. 16J. p. 67. Exercise 1 1. 1. 12. 2. 72-5. 3. 42. 4. 15. 6. 13-6. 6. 21, 23, 25. 7. 10 miles. 8. 40. 9. 11-9 in.. 101 in. 10. 4 men; 60p. 11. 5. 12. Son 8 years, father 32 yeara. 13. 30. 14. 80 at 18p, 40 at 12p. 16. 23 p. 70. Exercise 12. 1. 175. 2. 154. 3. 1131. 4. 856. 6. 0616. 6. 57?.' 7. 4|. 8. 204. 9. i. 10. 1390 (approx.). 11. 31-4. p. 74. Exercise 13. — # --# *'-V5 a r 825H IB-/* 112 A x 10 1 i.-ffl* *-«#*■ 1H (2) R =j; o-i. 8. (1) 7 - /R. '<?• INR - 9w -V r - i 10. A -1008; «=^p. p. 77. Exercise 14. 1. &». 2. 2a. -26 6 ' 6/) + o 4. 46. 6. 3a " 290 7 2(6 - a) '»■ *i± g' + 6» a-b' TEACH YOURSELF ALGEBRA 6- 196 13. 9. p. 12. 12a 4 + 3a - p. 82. 1. x = 3, y = 6. 3. x = 12, y = 3. 6. x = 3, y = 1. 7. x = 7, y = 4. 9. * = 6, y = 6. 11. * = 6, y = 10. 13. a = — 2, 6 = 3. 16. x = j, y = - 2. 17. a = - 7, 6 = 8. 19. * = 8, y = 12. 21. * =4, y = 2. 23. * =, Y, y «= _ 3. Exercise 15. p. 86. 10. 1. x = 12, y 3. 20, 7. 6. m = 2, 6 = - 3. y-9. 7. a = 05, 6 = 06. 9. 60 ft.; 36 ft. 11. Tie 49p; socks 24p. 2. * ~ 5. y = 8. 4. x = 8, y = 3. 6. * = 4, y = 6. 8. x = 10, y = 7. 10. * - 3. y = 10. 12. a = 5, 6 = - 9. 14. p = - 2, q = 4. 16. * =. 9. y = 13. 18. * «= 6, y = - 4. 20. * = 1-35, y = 2-7. 22. P = 1-8, Q = 032. 24. * - 12, y = 6. Exercise 16. 2. 16, 12. 4. 10, 0. Equation is y = 2x — 3 ; when * = 6. m = 20, b = 10. 8. a = - 1-36, 6 = 138. 10. £3-75: 2p. 12. u = 10,/= 6, 125 ft. p. 96. Exercise 17. 4. (1) 5. Tht 2. 52-5°. 3. Males: (1) 339. (2) 181. Females: (1 36-8. (2 202. 88-3 lb. (2) 130 lb. (3) 3 lb. The birth rate in 1931 could not be safely forecast as the graph is irregular; the actual birth rate was 15-8. 6. Yes; 251. 7. (1) The graph is approximately a straight line and ap- parently a law connects the load and effort. (2) 15-4 lb. approx. 8. 1370 tons; 1700 tons. 9. /111. ANSWERS p. 105. Exercise 18. 1. (1) 2. (2) - 2. 2. (1) - 1. 3. 1) - l (2) 4, 4. (1) 3. 5. (1) 4. (2) 6. 6. (1) - i. 7. a = 2, b = 3. Equation is y = 2x + 3. 8. * + 2y - 4. 291 i 2) I- 2) 2. (2) 1. Intercept is 3. C, (-2, 3); F, (3. 0); p. 108. Exercise 19. 1.^,(4,4); B. (4-6. 12); D. (4, -2); E. (- 1-4, - 34) ; G, (0, - 3). 2. (1-12, 1-12). 4. (2, 2). 6. They lie on a straight line parallel to OX. 6. They appear to lie on a straight line which passes through the origin. 7. For every point on it the * co-ordinate is + 3. p. 116. Exercise 20. 1. All lines pass through the origin with different slopes. 2. All lines are parallel with slope 45°. 3. All lines are parallel with gradient J. 4. All lines have the same intercept on the y axis, viz. 2. 6. Intercepts are : (1) 1-5 and 3. (2) 4 and —2. (3) 2 and 5. (4) 25 and - 2 6. (1) * = 4, y «= 3. (2) * = 3,y«l. 3) x = - 2, y = 2. 7. a -3; - 1. 8 6-1; 1. 9. a •=• 3, 6 = — 2. Equatior isy =» 3* — 2; — 2. 10. y = 2x + 3. 11. 3y — 2* -» 7. Gradient - 1 12. y — x = 2. Gradient *» 1. p. 121. Exercise 21. 1. ab + ay + bx + xy. 2. ce + cf+ de + df. 3. acxy + adx + bey + bd. 4. ab — ay — bx + xy. 5. ax — bx — ay + by. 6. ab + ay — bx — xy. 7. ab — ay + bx — xy. 8. a6 + 3a + 26 + 6. 9. aft - 3a - 26 + 6. 10. a6 + 3a - 26 - 6. 11. a6 - 3a + 26 - 6. 12. jt» -J- 12* -f- 35. 13. a«6' + 9a6 + 18. 14. x* + 13* + 30. 15. *• - 13* + 30. 16. *• + 7* - 30. 17. x* - Ix — 30. 18. p * — 4p — 96. 19. x a — 12*y + 32y«. 20. *• + 4*y — 32y». 21. x*- 4xy - 32y". 22. 2a« + 9a6 + 106". 23. 9*> - 27*y + 20y«. 24. 28*' + 16* + 2. 25. 6*» — 1L* + 3. 26. 9*« - 9* - 4. J9» 27. 29. 31. 33. 35. TEACH YOURSELF ALGEBRA 1 — y - 12y». 14*» + 2<dxy - I6y*. a + b. 12. (a) x*-y*. (c) ' 1 + *». (6) a' (d) r» 18*' - 27* - 5. 18a« - 67a6 + 356«. a + b — c. 5y — 5x •= 0-5. 88. 30. 88. 34. o» + 8. + 3x«<j + 3xa* + a*. p. 124. Exercise 22. 1. 3. 5. 7. 9. 11. 13. 16. 17. 19. 20. 21. 22. 23. 24. 25. 27. 29. 31. 33. 35. 37. 39. 2. x* - \x + 4. 4. a« — 6ab + 96*. 6. *» — 4xy + 4y*. 8. x*y> - &vy + 9. 10 12. 14. 16. *» + 4* + 4. a* 4- 6a6 + 96'. 4** + \xy + y*. a'6« + 20a6 + 100. 16*' + 40*y 4 2fl»* 25*V + 60*y + 36. 25** + 30*V + Oy 4 - y y* *• + 2xy + y* + 2* + 2y + 1. 1 — 2* + 4>- + **-4*y + 4y». a« + 6» + e' + 2a6 - 2ac - 26<;. ** + J>* + ** — 2*y + 2xz - 2yz. 4x* + 9y* ~ 25** + 12*;y — 20« - 30yj. 16a» + 46« + 1 - 16a6 - 8a + 44. 16*' - 40*y + 25y». 1 - 20*« + 100*«. 9*V — I2xy* + 4y*. S* xy + ?• 16" «■*-? + *. *» + 3**y + 3*j-« + y>. 20. a» + 6a' + 12a + 8. 28 p* + 3p*q + 3pq* + o». 30. Sx' + I2x*y +- 6xy* + y*. 32. 27a 3 - 27a 3 + 9a - 1. 34. 2*y. 36. 40*. 38. 4a(* + a) ; 256 sq. ft. 40. x* — 3x*y + 3xy* — y«. a» — 6a« + 12a - 8. p* - 3p'q + 3pg* - o>. ** — dx*y + 12xy' — 8y s 1 - 96 + 276' - 276 s . (1) 4a6; (2) - 4a6. 9y* + 18y + 9; 36. 2<*(* + y + 2a). p. 126. 1. a« - x*. 3. a» — 46». 5. 4*' - 1. 7. 1 — a*. 9. x* - y*. 11. 144**y« - 1. 13. tx + y)» - *». 15. (2a + 36)» - 1. 17. a» — 4(6 + c)». 19. x- - J. 21. (x + a)» - i. Exercise 23. 2. p' - qt. 4. 16*» - 9. 6. 1 - 36*». to*— L 9x*y s — 4. i*x* — 49. (a + *)« - y«. (x - 2yy - 36. 4*" - 9(y + *)«. x' y* 4 9" 8 10 12 14 16 IS 20 *;-< p. 128. 1. 6(* + 2). 3. 2y(2* + y). 6. 7xy(2xy - 1). 7. a(a — b + c). 9. a*(15a — 66 H 36»). 11. a6(a + b - c). 13. 7-4(13» + a»). 14. 186(18-6 + 1-4) ANSWERS Exercise 24. 2. a(36 + 2). 4. 2a(3« 26). 6. 16(1 - 2a»). 8. x{x* + 3x - 1). 10. 3a«(2a - 6c). 12. *f ^ - ^ 3J" 18-6 x 20 = 372. 9. (a — 6 11. (ax - I * + y). lb +e). (x + q . b + 6. P + 6. b)(x + a). Exercise 25. 2. 4. 6. 8. 10. 12. -y)(a- c\. - g\(* - h). a - 5)(6 - 6). 2a + 3)(6 - 5). x -b)(x + a). p. 132. Exercise 26. 1. x + 3. 5. 7. 9. 11. 13. 15. 17. 19. 21. 23. 25. 27. 29. >+ 1). x + Z)(x + 2 . * + 6)(* + 1). * - 10y)(* - 2y). ixy + 6)(*y + 9). u, _ nly - 12). fx - 2)[x + 1). * + 3y)(x - 2y). b - 3)(6 + 1). x + ieW - 3). x - Uy)(x + lOy). a - 4)(a + 3). U> - 36)(p + 2). II - 10*)(1 + 2x). P + 9)U>~ 6). a - 126)(a - 36). a6 - 16)(a6 - 3). (* - 1y)[x - ay), (x + 2)(x - 1). 6. (x - 3y)(x + 2y). 1. (6 + 3) (6 - 1). \x - 16) (x + 3). a - 12j (a + 1). \p + mp - si- ft - 6*)(1 - 4x). {xy - ll)(xy + 8). Exercise 27. £ + 2). 2x - 6). \x + 2). 2x - 1). 3x - 1). i±i ^ x i - 0(3* 2. (4* - 3)(3* - 2) 4. (9* - 2 ) * + 5). 6. (3* - 1 8. (3* + 1 10 (5* — 1 12. (4* + 1 14. (2a + 1 16. (56 + 2) 18. (2y — 6) 20. \lc + 2) ^ + 5). x- 1). 2*+ 1) I* - 1). 3x + 2). a- 1). 26 - 1). iy + 3). 2c - 3). 293 294 TEACH YOURSELF ALGEBRA p. 137. Exercise 28. [p 4- ?)*. (3* + 1)' 5. (* 4- J)'. 2. 4. a l* - ay) 1 . (4* - By)*. 7 9 11 13 15 17. 19. 21. 22. 24. 26. 28. 30. 32. 34. 36. 38. : [a 4- 6 + 2)«. * 4- 10)(* _ 10). (2* 4- 3y)(2* - 3y). 11*4- 6v)(ll* - «y). !5 + 4a)(5 - 4a). 2(2a 4- 56) (2a - 56). 5(* + 3y)(* - 3y). [x 4- 2y 4- 4«)(* + 2y - 4*). l G+J)' .1 + W(i - ?»• (a + * — 2y)(a — x + 2y). iab. 2100. 252. 150. 68. 630. 20 60. 8. (x — y - 5)'. 10. (ab 4- 6)(a6 - 5). 12. (5a + 46)(5a - 46). 14. (12/> 4- 13?)(12£ - 13o). 16. (1 + 15.t)(l - 15*). 18. 3(* 4- 5)(* - 5). >. (a 4- 6 4- e)(a + b - c). 23. {x+y + t )(x-y- z). 25. (* - 16) (* - 1). 27. 3000. 29. 880. 31. 15. 33 264. 35 38-4. 37. 448. 139. Exercise 29. 9. 10. (* + £)(** — ex + c*) 2. (1 + 2a)(l - 2a + 4a«). 4. (2 + 3e)(4 - 6t 4- 9c«). 6. (m — 5n){m* 4- 5>nn + 25>i»). (*y4-.l)(*Y -f +i). \x + y){x*~ xy + y'h \x~ y)(x* + xy + v»)- [y - a)ty* + ay + a«). (* - 4)(*« + 4* 4- 16). (H - l){li> + It + l). p. 141. Exercise 30. ■Ad* x — a{a 4 - 36) bW+ 26)' 8. 4. 6. 8. 10. a*- a* + 36 S6Ta-T)' a +6 a x - 5 x + 5' * +y x* + xy + y* ANSWERS 295 p. 142. Exercise 31. l. *■% (F+~n- a(« 4- 26) a — 6 7. 1. 9, 2 j b - 4 > . a. p. 145. I. 2x — y x(x - y)' „ 13a — 56 6 - E$?~=~6*)' _ 3*' - 20* + 34 o. 7. (* - 3)» 7 - 3a (1 - a)«- 2y - *' - 2v.v *(* 4- ty) ' 11. <(« ~ b) (1 - a/)(l - «)" 2* - 1 ,3 - |r=jjt is. ;? = n P + q 17 /? Sl£l 17 - " ■ 2i?;+ 3/?;- 21. 2i>Q 23. 1-47. 2. y(y 4 1). *• a - 3- (*4- l)(*-3) 2(* - I) - 8. 10, £±1.; 2-25. Exercise 32. 2 - » + *> , * (* - 1)(* - 2)- *» 4- 2* - 2 *(* + 2) • - fi*« - 17* 6(* 4- D(* - 2)' 12* 6. 8 - **-■ „il06 -3a) IW - 86(8a — W (*-a) -H(y-6) (a + 60 (* + yl) ' — 2y* 12. 14. (* - y)'(* 4- y)' 16. R ,-HL 9-; 5p-q 22. flg^p. 2 9 6 p. 147. I. 3. 4. -». 7. 7. 10. 6. 13. 8. TEACH YOURSELF ALGEBRA Exercise 33. Ik 14. V RE R + r 3. 2-5. 6. 8. 0. 4. 12. 6. p. 163. Exercise 34. Answers involving decimals are mostly approximate. 1. 2. :i. 4. 6. 6. 7. 8. 9. 10. (1) 5-3. (2) ± 2-65. (1) ± 1-8. (2) ± 2-83. (1) ± 2-45. 2 ± 2-24. (1 ± 1-41. (2) ± 2-45. (1) 3-73 or 027. (2) (3 4-24 or — 0-24. (1) - 3-41 or - 0-59. (3) 0-83 or - 4-83. Min. value is — 4 when x = 3. (1) 6 or 1. (2) Min. value is — 2 when x — 2. (1) 4-46 or - 045. (2) 3) 3) 3) ± 1-87. ± 346. ± 141. - 4 or — 2. (2) - 3 or — 1. 61 6 or - 016. 5 or — 1. Min. value is — 1*1 approx. when x = $. On substituting x — f in the function, this equals — 1J. (1) 2 or J. (2) 2-69 or — 019. Max. value is 2-25 when x =» — 0*6. p. 169. Exercise 35. 1. ±2. 2. ±i- 3. ± 12. 4. 1. - 3. 5. 8, - 2. 6. 1, - 11. 7- i. - i- 8. 10-75, 1-25. 9. 8, 2. 10. 3, - 4. 11. 5. - 3. 12. 4, - 7. 13. 8. - 4. 14. 2, S. 15. 9, - I. 17. 1-434, 0- 16 32"5. - 0-92. 232. 18. 4 and — 1-5. 19. 6. - J. 20. 2-732, - 0-732. 21. 1314, - 114. 22. 6-69, - 7-69. p. 171. Exercise 36. 1. 0, 3. 2. i', - 5. 3. 2. 2. 4. 1, 2. 6. 1, - 4. 6. 3, - 7. 7. i. - V- 8. •t. 5. 9. 2, - 3. ANSWERS 10. 5, - 7. 11. - 3, - 10. 12. 11, - 6. 13. 9. - 5. 14. a. 4. 15. ft. - 1. 18. J, - |. 16. 1, i. 17. J, - j. 19. 2, 4. 5. 20. }. - 3, - 2. 21. 1, 4. - 2. 22. J, 10, - 5. 23. a. b. 24. |, - d. p. 175. Exercise 37. Answers are approximate. 1. 0-303, - 3-303. 2. 4-661, 0-438. 3. 1-48, - 108. 4. 1-43, 0-23. 5. 2-85. - 0-35. 6. 1-744, - 0-344. 7. 0-535. - 6-535. 8. 3-441, — 0-775. 9. 0-427, - 2-927. 10. 0-803. - 2-803. 11. 1081, - 1-481. 12. - 0-130, - 0-770. 13. 2591, - 0-257. 14. 3-68, - 0-43. 15. 9-75, - 1-75. 16. 2-13, - 6-62. 17. 6-123. - 2123. 297 p. 177. Exercise 38. 1. ft or *. 2. 9 in., 15 in. 3. 6-63 or - 1-63. 4. 2137. 2737. 5. 6-65. Negative root has no meaning for the problem. 6. 79-78. Second root is inadmissible. 7. 12-5, - 10. 8. 8. Negative root is inadmissible. 9. 16 in., 20 in. 10. 16 sq. yds.. 25 sq. yds. 11. 7, - 1. 12. 12. Negative root is inadmissible. 13. 108 sq. yds. 14. 78-4 ft. 15. 8 and 12. or — 12 and — 8. 10. 12. p. 180. Exercise 39. 4; * — — 6, y - - 7. 2; x — — 26, y - 33. i; * - - A. V - ff • 6i; x -31,y- - 1\. 1. 2. 3. 4. 6. 6. 7. 8. 9. 10. 11. O.y s.y z.y *»* -3,y l.y - 4; 4, y = 3 ; — 14. y * 7, y ■= 4 ; 6, y = 4; 2,y = 4; 2; x = — 2,y - 1. * - f> y - »■ ** 14' J 14 ' 11; *-j,v--4. »-t,y-il. * - - A. y - - W- * - t.y - - I 298 TEACH YOURSELF ALGEBRA p. 182. Exercise 40. 1. x = 3, y — 5; x = 5, y = 3. 2. * - 7, y = 2; * = 2, y = 7. 3. * — 8, y = 3; x = — 3, y = — 8. 4. * = ± 4, y = ± 1 ; * = ± 1, y = ± 4. 6. * = 6, y = 2 ; * = — 4, y = — 3. 6. * — 4, y = 1 ; * = — 1, y - — 4. 7. x — 9, y ■= 2; * = 2, y = 9; * = — 2. y = x = — 9, y = — 2. 8. * «• 7, y = 1; * = 1, y = 7. 9. x — 6, y = 3 ; * = — 6, y = — 3 ; * = 3, y x = - 3, y = - 6. 10. * = 3, y = 1; * = 1-5. y = 2. ANSWERS «99 - 9; 6; p. 186. Exercise 41. 1. Co 9 . 2. a". 3. i«'- 4. 2'° = 1024. 6. 4* = 4096. 6. 6a 8 6'. 7. *«y»V. 8. x 1 '. 9. ««-. 10. a t '~'b'. 11. a*. 12. 3--H + 4. 13. a'. 14. J*», 15. 3a». 16. 2« = 04. 17. 3» = 27. 18. *• 19. - 6a 3 b. 20. a'. 21. Ax>y. 22. a'. 23. *. 24. «»-'«. 25. a". 26. a*. 27. Sa& "9" - 28 *' 28. p. 29. 2* = 04. 30. 3« = 72 31. *>°. 32. *>° 33. a". 34. 4a«6». » "r- 30. a". 37. x*". 38. 27a". 39. o«. 40. **. 41. x». 42. 3a'. 43. a«6. AA ** 44. -i. y* 40. 3a' 46. x". p. 193. Exercise 42. 1. (1) ys- (2) ^8. (3) &a. (4) V2"- (5) ^3. («: y&. 2. (1) i/1000. (2) ^T6. (3) VaK (4) lOVlO. (6) 4v/2. (6) io^is: 3. (1) 2^2. (2) 4-*'2. (3) v «'• (4) V» '"-a*V5T (5) v^a* -a^*. (8) /a. 4. (1) 8. (2) 32. (4) 1000. (5) 316. 6. J, J. 1. 3. 3\/3. 9. 9\/3. 6. (1) 5050. (2) 27. (4) all. (5) 2. 7 - *■ «*' 71- 3 ' **' looo- (3) 27. (6) i- 8- (1) t3 . 9. (1) 4. && 10. (1) 4. (4) *• 11. 5656. 12. (1) •x-'al 1 15625- 1 13. (1) a». 14. (1 3-52 x 10». 4 3-7 X 10*. p. 202 1. 1. 2. (1) (2) 3) 4) ffl CM (5) x. (2) 12S. (5) 16. ;2> v- ») 8. (2) lOOyfO =310-2. (5) x". (2) 1. (2) 1-633 X 10'. (5) 2-52 X 10'. Exercise 43. 3. 3, 4, 2. 0, 5, 1. 3, 0. 2. 0-6990, 1-6990. 26990. 46990. 0-6721, 2-6721. 4-6721. 1-7226, 0-7226, 27226. 2-9767. 0-9767. 49767. 0-7588. 1-9842, 38433. 446-7, 44670, 4467. (3) *». (6) i*». (3) 1000. (6) 3. [3) i- !o) A- (3) j,. (6) x*". (3) 0. '3) 2-2 x 10>. 2) 87-70, 8770, 8770. p. 204. 1. 3440. 5075. 1-589. 13-56. 3137. 104-6. 1-359. 2-786. 4. 7. 10. 13. 16. 19. 22. 71.400. (4) 2628, 5-229, 1140 Exercise 44. 2. 276-4. 3. 1397. 6. 2-396. 6. 6-997. 8. 222-8. 9. 14-22. 11 851 3. 12. 2650. 14. 728-5. 15. 2172. 17 1 -656. 18. 1436. 20. 1-695. 21. 2-321. 23. 5002. 24. 1-546. 300 TEACH YOURSELF ALGEBRA p. 206. Exercise 45. 1. (1) 0-4470, 1-4470, 5-4470. (2) 0-6298, 1-6298, 36298. (3) 3-9904, 4-9904, T -9904. (4) 2-8097, 1-8097, 3-8097. 2. 3. 2-7771. 5-9673. 0-2159. 0-0004402. p. 209. 1. 2. 3. B. 4-6037. 2-5926. 2-4814. 1-7464. 1-3673. 2-6856. T-1463. 1-7399. 2-5598. p. 211. 1. 15-42. 4. 6-699. 7. 004903. 10. 0-8414. 13. 007115. 16. 3-558. 19. 01429. 22. 003129. 25. 28. 31. 34. 37. 40. 1-663. 896-6. 2-89. 1-441. 530-7. 1200. p. 218. 1. (a) 100a: b. 3. (1) f. 4. (a) 35. c(b + c) 6. a ax 4-6749. 1-7538. 0007453. 0-5940. Exercise 2) 2-7126. 2) 0-8263. 46. g m 1-8368. 1-1968. T-07155. 20254. T-7127. 1-7206. Exercise 47. 2. 6. 8. 11. 14. 17. 20. 23. 26. 29. 32. 35. 38. 41. 0*8886. 0-6116. 0-lti00. 01226. 1-826. 5-471. 9-399. 1-022. 3-940. 102-2. 1035. 32-2. 4-923 ft. 0-8268. Exercise 48. (6) 2 240? + r (2) 25 : 64. S-90U. 2-9023. 003070. 2-482 x 10-'. (3) 1-6697 3-8910. 4-7913. 1-7754. 0-6619. 2-7726. 3-8973. 3. 6. 9. 12. 16. 18. 21 24. 27. 30. 33. 36. 39. 001529. 003239. 85-23. 1197. 1-467. 01014. 0-3609. 62-84. 9-527. 516-4. 13-05. (<*) 3; (1) 1-5261, (2) 3-2193. (b) 3. 2. H . (3) 5 : 12. (0l 8, pA qA 7. _==_ lbs., . 7- lbs. * + *"•_* +y 10. (1) bVac. (2) 4ab. I + ?' 8 - 'a p + <r 9. 17. (3) (a + b) VaF. 11 7. 12. \. 14. (a) 450. (b) ANSWERS 13. (1) i. (la 15. x = ^a*S, y - 4 ab- 5b- (2) ?• W 35- 3oi (3) J. p. 226. Exercise 49. 1. (a) No. The runner's rate differs at various parts of the race. (6) Yes. (c) No. (a") Yes. (e) No. The connecting law is as explained in § ^04. 2 6; 46ff. 3. 375; 336. 4. y = V*i 4» . 6. s = 32/; 8-96 ft. 6. £ — A W ; 3-73 ins. 7. y =- i* + 6. 8. £ = 0-06W + 0-2. 9. F •= Q42W - 00. p. 234. Exercise 50. 1. y = jj*»; Ml. 2. y 3. y values 12, 3; * value 6. 5. 384; 2^4. 4. v 7. * values 2, 15: y value 10. 9. 32-U miles (appro*.). 11. d = 2^/g ; 40-5. 13. 7-21 ft. ■ Vx ; 8-76. 30 i - «• y = x -■ I'"- 1920 10. 95 lbs. (appro*.). 12 4-08 ft. 14. v m Hx>'. 8. /• = p. 239. I. (a) y a. xx. (b) y « p. Vx (e) y a — . (d) V oc Ar». (/> v . 4. 4. 10A. 0-24/£» 7. H = 8. 31Jcwt. ft Exercise 51. .-. v = kxz. kx ..y= jT . kVx .-. y = ; • .-. v = Mr*. ,. W = ft f . ..y-A.^j. 3. y = 3.vj- ; 8. 6. 8 tons. 6. 2 018 ohms. (1) 24.000 units. (2) 180 sees. 302 TEACH YOURSELF ALGEBRA p. 246. Exercise 52. 1. y = e*« + 3. 3. R = 001K» + 5. 5. y = 0-54 + 0-24**. 7. y = 0-25* 3 . 9. Q = 0-7//°". 2. >> = 0-42* 8 + 7-5. 4. tf = 1-08 J" -I »20. 8. y = 3*». 10. y = 3-65*». p. 252. 1. (1) (4) 2. (1) (4) (7) (10) (12) 3. (1) (4) (7) 4. (1) (4) (7) (10) 6. (1) t (4) Exercise 53. VI50. (2) V288. V208. (5) v/4^5. 20 V2. (2) 8 V5. 50 V2._ (5) 5Vl6._ 2«6V6a ; _ (8) 3x t yV2yt. lOp VlO?. (11) a{a - 6) Vb(a - b). V500. VWy. (3) (6) (3) 0V3. (6) 60V2._ (9) 5abcVSabT. 5y(x — 2y) Vxy. 3 V5. (2) 7V2. lOVTO. (5) (o + 6) Va — b. 2V2 - I. 23 + 4V15. 38. (3) 16 VS. - V2. (6) 7V2 + V J4". {8)_ 3(* - 2y) V2(x + 2y). (2) 2-3 Ve. (5) 67 - 10V14. (8) 301 + 20 V3. (7) -V V5 5 \/2 8 2V35 7 (2) "V" (5) ^ (8) H°- 0) ^ (6) 2 (10) (12) (U) (16) (18) p. 259. I. (1) \ 3 \ 4) (5) 21. 2VS 3 VlO 30 ' Ve 3 3(V7- VS ) 6 5(2 V7 + 3 VS). 6 — VSiT 2V2. 4(0 + Ve) 31 Exercise 54. 12-6, 15. 17-5. (2) 0, - 4, - 8. (a - 36), (a - 56), (a - 76). 6-6, 7-9, 92. (x + 2y). {x + 3y), (x + 4y). (3) 3- 2V2. (6) 2. (9) a - 25. (») 3( 2 VS + V2) '"' " 10 (13) 5 + 2 VS. »« 2 -4^- (17) - V3. 3V7 14_ 6V6 5 ' V5+ 1 ANSWERS 3°3 2. 12, 16-5. 3. 4 + 4/>. 4. x + 3n - 1. 6. 348. 6. 29. 7. 2. 8. 19. 9. (a) 217-5. (6) 16. (c) 0. (rf) 304i. 10. 12. 11. 6 or 8. 12. 120; 1860. 13. £40-05; £5343-75. 14. £34. p. 263. Exercise 55. 1. (a) 62-5, 156-25, 390-625. (6) }, A, A- \c) - 64. 81, - **a. id) 3 X 10-*, 3 x'10- 8 , 3 x 10 V (e) 0-010126, 000151875. 0-0002278 (approx.). 2. 320 3 l,B . 4. 1-61051. 5l 6-001215. 6. {a) ar 2 "- 1 , ar«"; — ar*"- 1 , (w*-. 7. 105. 8. 104. 9. (1) 3-873; (2) 3-888 (both approx.). 10. 7, 9-8. 11. £146410. 12. £162,901-25. 13. I 04. 14. 10. 20. 40. p. 266. 1. la) 94-5. 2- W- 6. 2168. 8. 93-8 approx. p. 271. 1- (")!- Exercise 56. (6) 1993. 3. m- 6 765. Exercise 57. (0 H- 4. 16-32. 7 27, 40JJ. (a) i8- Series (e) is the sum of (a) and (6). (&) i- (c) 1. (b) A- (O J S- the sum of (a) and (6). It represents 0-161616 .... a recurring decimal. 3. |. This series equals the scries (6) in question (1) increased by unity and then diminished bv series (a). 4. In f{l — (J)-}, (J)- — > o as « — >'oo . .-. the limit is jj. 6. (<*) Y- 0) ¥• W V 7. (a) 26. (6) 2(V2 + 1). 8. f. 9. 112 ft. 10. £76,000. p. 276. Exercise 58. Most of the answers are approximate. 1. £1180. 2. £5500. 4. £2430. 5. £3431. 7. £1067. 8. £240,000. 3. £486. 6. £5000. LOGARITHMS of numbers 100 to 549 Proportional Parts LOGARITHMS of numbers 550 to 999 • 1 2 S 4 • 1 7 • t 1 2 3 4 1* 7 • * 10 0000 0043 0086 0128 0170 0212 0253 0294 0334 0374 4 S 12 17 21 25 29 13 37 II 0414 0453 0492 0531 0569 0607 0645 0682 0719 0755 4 8 11 15 19 23 26 30 14 12 0792 0628 0864 0899 0934 0969 1004 1038 1072 1106 3 7 10 14 17 21 24 28 11 13 1139 1173 1206 1239 1271 1303 1335 1367 1399 1430 3 6 10 13 16 19 23 26 29 M MB 1492 1523 1553 1584 1614 1644 1671 1703 1732 3 6 9 12 15 18 21 24 27 IS 1761 1790 1818 1847 1875 1903 1931 1959 1987 2014 3 6 8 11 14 17 20 22 25 It 2041 2068 2095 2122 2143 2175 2201 2227 2253 2279 3 5 8 11 11 16 18 21 24 17 2304 2330 2355 2380 2405 2430 2455 2480 2504 2529 2 5 7 10 12 15 17 20 22 IS 2SS3 2577 2601 2625 2648 2672 2695 2718 2742 2765 2 5 7 9 12 14 16 19 21 It 2788 2810 2833 2856 2878 2900 2923 2945 2967 2989 2 4 7 9 11 13 16 18 20 20 1010 3032 3054 3075 3096 3118 3139 3160 3181 1201 2 4 6 8 11 13 15 17 1* 21 1222 3243 3263 3284 3304 3324 3345 3365 3385 3404 2 4 6 8 10 12 14 16 18 22 1424 3444 3464 3483 3502 3522 3541 3560 3579 3598 2 4 6 8 10 12 14 15 17 21 1617 3636 3655 3674 3692 3711 3729 3747 3766 3784 2 4 1 7 911 13 15 17 24 3801 3820 3838 3856 3874 3892 3909 3927 3945 3962 2 4 5 7 9 11 12 14 14 21 1979 1997 4014 4031 4048 4065 4082 4099 4116 4133 2 3 5 7 » 10 12 14 IS 24 4150 4166 4183 4200 4216 4232 4249 4265 4281 4298 2 1 S 7 8 10 11 11 IS 27 4114 4330 4346 4362 4378 4393 4409 4425 4440 4456 2 3 5 6 8 9 11 11 14 28 4472 4487 4502 4518 4533 4548 4564 4579 4594 4609 1 3 5 6 8 9 11 12 14 It 4624 4639 4654 4669 4683 4698 4713 4728 4742 4757 I 1 4 6 7* 10 12 11 M 4771 4786 4800 4814 4829 4843 4857 4871 4886 4900 1 1 4 4 7* 10 11 13 11 4914 4928 4942 4955 4969 4983 4997 SOU 5024 5038 1 3 4 5 7 10 U 12 12 5051 5065 5079 5092 5105 S119 5132 5145 5159 5172 1 1 4 S 7 8 9 11 12 11 5185 5198 5211 5224 S237 5250 5263 5276 5289 5302 1 J 4 S 6 B * 10 12 M 5315 5328 5340 5353 5366 5378 5391 5403 5416 S428 1 3 4 S 6 8 9 10 11 IS S441 S453 5465 5478 5490 5502 5514 5527 5539 5551 1 1 4 16 7 9 10 11 u S563 5575 5587 5599 5611 5623 5635 5647 5658 5670 1 2 4 S 6 7 a io ii 17 5682 5694 5705 5717 5729 5740 5752 5763 5775 5786 1 2 3 S 6 7 8 9 10 18 5798 5809 5821 5832 5843 5855 5866 5877 5888 5899 ! 2 3 S 6 7 8 * 10 1* 5911 5922 5933 5944 5955 5966 5977 5988 5999 6010 1 2 3 4 S 7 8 t 10 40 6021 6031 6042 6053 6064 6075 608J 6096 6107 6117 1 2 3 4 S 6 7 9 10 41 6128 6138 6149 6160 6170 6180 6191 6201 6212 6222 1 2 3 4 S 6 7 8* 42 6232 6243 6253 6263 6274 6284 6294 6304 6314 6325 1 2 3 4 I 6 7 8* 41 6335 6345 6355 6365 6375 6385 6395 6405 6415 6425 1 2 3 4 S 6 7 8* 44 6435 6444 6454 6464 6474 6484 6493 6503 6513 6522 1 2 3 4 S 6 7 8* 4S 6512 6542 6551 6561 6571 6580 6590 6599 6609 6618 1 2 3 4 16 7 8* 44 6628 6637 6646 6656 6665 6675 6684 6693 6702 6712 1 2 3 4 S 6 7 7 8 47 6721 6730 6739 6749 6758 6767 6776 678S 6794 6803 1 2 3 4 S 5 6 7 S 48 6812 6821 6830 6839 6848 6857 6866 6875 6884 6893 1 2 3 4 4 S 6 7 6 4t 6902 6911 6920 6928 6937 6946 6955 6964 6972 6981 1 2 I 4 4 S 6 7 • SO 6990 6998 7007 7016 7024 7033 7042 7050 7059 7067 1 2 3 1 4 5 6 7 8 SI 7076 7084 7093 7101 7110 7118 7126 7135 7143 7152 1 2 3 J 4 S 6 7 8 a 7160 7168 7177 7185 7193 7202 7210 7218 7226 7235 1 2 2 1 4 S 6 7 7 si 7241 7251 7259 7267 7275 7284 7292 7300 7308 7316 1 2 2 1 4 S 6 6 T S4 7124 7332 7340 7348 7356 7364 7372 7380 7388 7396 1 2 2 1 4 S 6 6 7 ' 2 S 4 » 6 T a | r 1 2 1 4 S 4 7 8 * Prop. irtlonal P im 1 2 1 4 S 6 7 8 » 1 2 3 4 S 4 7 8* SS 7404 7412 7419 7427 7435 7443 7451 7459 7466 7474 1 2 2 56 7482 7490 7497 7505 7513 7520 7528 7536 7543 7551 1 2 2 S7 7559 7566 7574 7582 7589 7597 7604 7612 7619 7627 1 2 2 58 7634 7642 7649 76S7 7664 7672 7679 7686 7694 7701 1 1 2 S* 7709 7716 7723 7731 7738 7745 7752 7760 7767 7774 1 1 2 60 7782 7789 7796 7803 7810 7818 7825 7832 7839 7846 1 1 2 61 7853 7860 7868 7875 7882 7889 7896 7903 7910 7917 1 1 2 62 7924 7931 7938 7945 7952 7959 7966 7973 7980 7987 1 1 2 61 7991 8000 8007 8014 8021 8028 8035 8041 8048 8055 1 1 2 64 8062 8069 8075 8082 8089 8096 8102 8109 8116 8122 1 1 2 « 8129 8136 8142 8149 8156 8162 8169 3176 8182 8189 1 1 2 66 8195 8202 8209 8215 8222 8228 8233 8241 8248 8254 1 1 2 67 8261 8267 8274 8280 8287 8293 8299 8306 8312 8319 1 1 2 68 8325 8331 8338 8344 8351 8357 8363 8370 8376 8382 1 1 2 69 8388 8395 8401 8407 8414 8420 8426 8432 8439 8445 1 1 2 78 8451 8457 8463 8470 8476 8482 8488 8494 8500 8506 I 1 2 71 8511 8519 8525 8531 8537 8543 8549 8555 8561 8567 1 1 2 72 8571 8579 8585 8591 8597 8603 8609 8615 8621 8627 1 1 2 71 8633 8639 8645 8651 8657 8663 8669 8675 8681 8686 1 1 2 74 8692 8698 8704 8710 8716 8722 8727 8733 8739 8745 1 1 2 7S 8751 8756 8762 8768 8774 8779 8785 8791 8797 8802 1 1 2 74 8808 8814 8820 8825 8831 8837 8842 8348 8854 8859 1 1 2 77 8865 8871 8876 8882 8887 8893 8899 8904 8910 8915 1 1 2 76 8921 8927 8932 8938 8943 8949 8954 8960 8965 8971 1 1 2 7* 8976 8982 8987 8993 8998 9004 9009 9015 9020 9025 1 1 2 88 9031 9036 9042 9047 9053 9058 9063 9069 9074 9079 1 1 2 SI 9085 9090 9096 9101 9106 9112 9117 9122 9128 9133 1 1 2 82 9138 9143 9149 9154 9159 9165 9170 9175 9180 9186 1 1 2 81 9191 9196 9201 9206 9212 9217 9222 9227 9232 9238 1 1 2 84 9243 9248 9253 9258 9263 9269 9274 9279 9284 9289 1 1 2 81 9294 9299 9304 9309 9315 9320 9325 9330 9335 9340 1 1 2 86 9345 9350 9355 9360 9365 9370 9375 9380 9385 9390 1 1 2 •7 939S 9400 9405 9410 9415 9420 9425 9430 9435 9440 I 1 U 9445 9450 9455 9460 9465 9469 9474 9479 9484 9489 1 1 89 9494 9499 9504 9509 9513 9518 9523 9528 9533 9538 1 1 90 9542 9547 9552 9557 9562 9566 9571 9576 9581 9386 1 1 »l 9590 959S 9600 9605 9609 9614 9619 9624 9628 9633 1 1 92 9638 9643 9647 9652 9657 9661 9666 9671 9675 9680 1 1 91 9685 9689 9694 9699 9703 9708 9711 9717 9722 9727 1 1 *4 9711 9736 9741 9745 9750 9754 9759 9764 9768 9773 1 1 *l 9777 9782 9786 9791 9795 9800 9805 9809 9814 9818 1 1 H 9821 9827 9832 9836 9841 9845 9850 9854 9859 9863 1 1 »7 9868 9872 9877 9881 9886 9890 9894 9899 9903 9908 1 1 M 9912 9917 9921 9926 9930 9934 9939 9943 9948 9952 1 1 »» 9956 9961 9965 9969 9974 9978 9983 9987 9991 9996 1 1 1 2 3 4 8 6 7 8 9 1 2 1 4 8 4 7 8* ANTI-LOGARITHMS A NTI LOC jAK IHT b Proportion*! 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